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Found 85 results

  1. ingerNorway

    Checkbox-1/0-into database

    Hi I wondering what i did wrong here: I wont to register 1 or 0 into database. Checkbox file: <?php include("db-tilkobling.php"); $sqlSetning="SELECT * FROM Kvarterbestilt ORDER BY Møtt;"; $sqlResultat=mysqli_query($db,$sqlSetning) or die ("Ikke mulig å hente data fra databasen"); $antallRader=mysqli_num_rows($sqlResultat); print("<input type='checkbox' name='møtt' value='$møtt'>"); ?> Register-møtt.php <?php include("start.html"); ?> <br/> <p>Registrer Møtt/Ikke møtt ved å fylle ut skjemaet under</p> <form method="post" action="" id="registrerAnsattSkjema" name="registrerAnsattSkjema"> <fieldset> <legend>Registrer møtt / ikke møttt</legend> <label for="personnummer">Personnummer</label> <?php include("listeboks-personnummer.php"); ?><br/> <label for="timenummer">Timenummer</label> <?php include("listeboks-timenummer.php"); ?><br/> Sjekk av denne boksen hvis pasient har møtt til time:<br/> <?php include("checkbox-møtt.php"); ?><br/> <input type="submit" name="registrerMottKnapp" id="registrerMottKnapp" value="Registrer møtt/ikke"> <input type="reset" name="nullstill" id="nullstill" value="nullstill"> </fieldset> </form><br/> <?php /* include("valider-mott.php");*/ $registrerMottKnapp=$_POST ["registrerMottKnapp"]; if ($registrerMottKnapp) { $personnummer=$_POST["Personnummer"]; $timenummer=$_POST["Timenummer"]; $møtt=$_POST["Møtt"]; } if ($møtt == '1') { $query = mysql_query("INSERT INTO Kvarterbestilt(Møtt) VALUES('1')"); } /* $lovligFornavn=validerFornavn ($fornavn); $lovligFornavn2=validerFornavn2 ($fornavn); $lovligFornavn3=validerFornavn3 ($fornavn); $lovligEtternavn=validerEtternavn ($etternavn); $lovligEtternavn2=validerEtternavn2 ($etternavn); $lovligEtternavn3=validerEtternavn3 ($etternavn); $lovligYrke=validerYrke ($yrke); $lovligYrke2=validerYrke2 ($yrke); $lovligYrke3=validerYrke3 ($yrke); if (!$lovligFornavn) { print("Fornavn er ikke fylt ut! <br/>"); } else if (!$lovligFornavn2) { print("Bare bokstaver og mellomrom er tillat; $fornavn<br/>"); } else if (!$lovligFornavn3) { print("Ingen tall tillat; $fornavn<br/>"); } else if (!$lovligEtternavn) { print("Etternavn er ikke fylt ut!<br/>"); } else if (!$lovligEtternavn2) { print("Bare bokstaver og mellomrom er tillat; $etternavn<br/>"); } else if (!$lovligEtternavn3) { print("Ingen tall tillat; $etternavn<br/>"); } else if (!$lovligYrke) { print("Yrke er ikke fylt ut!<br/>"); } else if (!$lovligYrke2) { print("Bare bokstaver og mellomrom er tillat; $yrke<br/>"); } else if (!$lovligYrke3) { print("Ingen tall tillat; $yrke<br/>"); } */ /* include("db-tilkobling.php"); $check=mysqli_query($db,"SELECT * FROM Kvarterbestilt WHERE Personnummer='$personnummer' AND Timenummer='$timenummer'"); $checkrows=mysqli_num_rows($check); if($checkrows>0){echo "Ansatt eksiterer fra før";} else { $sqlSetning="INSERT INTO Kvarterbestilt(Møtt) WHERE Personnummer='$personnummer' VALUES ('$møtt');"; mysqli_query ($db,$sqlSetning) or die ("Ikke mulig Ã¥ registrere i db"); print ("Det er registrert at pasienten har $møtt til timen $timenummer"); } } */ include("slutt.html"); ?>
  2. 1. HTML FORM #for user to enter the data <html> <title>reg</title> <style type="text/css"> body { background-color: rgb(200,200,200); color: white; padding: 20px; font-family: Arial, Verdana, sans-serif;} h4 { background-color: DarkCyan; padding: inherit;} h3 { background-color: #ee3e80; padding: inherit;} p { background-color: white; color: rgb(100,100,90); padding: inherit;} </style> <form method="POST" action="login_back.php" enctype="multipart/form-data"></br> &nbsp<font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username"> </br></br> &nbsp<font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/> </br></br> &nbsp<font color="DarkCyan"> Last name:</font><input type="text" name="lastname"/> </br></br> &nbsp<font color="DarkCyan"> File: <input type="file" name="image"></font> </br></br> <input type="submit" value="Save and Proceed"> </form> </html> ---------- 2 STORING IN DATABASE #backend processing to store and retrieve data from db <?php error_reporting(0); #echo "<body style='background-color:rgb(200,200,200)'>"; session_start(); #if( isset($_POST['username']) && isset($_FILES['image']) ) #{ $_SESSION['username']=$_POST['username']; $_SESSION['firstname']=$_POST['firstname']; $lastname=$_POST['lastname']; $file=$_FILES['image']['tmp_name']; $image_size=getimagesize($_FILES['image']['tmp_name']); if(!isset($file)) echo"please select an image"; else { #$image=$_FILES['image']['tmp_image']; //grabing the file content $image_name=$_FILES['image']['name']; //grabing image name $image_size=getimagesize($_FILES['image']['tmp_name']); //getting image size } echo "</br>"; #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); #checking the available username $query = mysql_query("SELECT * FROM userdata WHERE username = '" . $_SESSION['username'] . "'" ); $ans=mysql_num_rows($query); if ($ans > 0) { echo "Username already in use please try another."; } else if($image_size==FALSE) { echo"That's not an image."; } else { #Insert data into mysql #1.Inserting user name & image into db $sql="INSERT INTO userdata(username, firstname, lastname, image)VALUES('" . $_SESSION['username'] . "', '" . $_SESSION['firstname'] . "', '$lastname','$image')"; $result1=mysql_query($sql); if($result1) { echo "</br>"; echo "Registration successful"; echo "</br>"; //displaying image $lastid=mysql_insert_id();//get the id of the last record echo "uploaded image is :"; echo "<img src='get.php?id=".$lastid."'>"; > this command has some mistake }#if insertion into db successful else { echo "Problem in database operation"; } }# else block of unique username n img }#end of isset ?> ---------- 3. GET.PHP #additional file that retrieve image from database <?php #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); if(isset($_REQUEST['id']) ) > this block of code is not running { $mid=(int)($_REQUEST['id']); $image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error()); $image=mysql_fetch_assoc($image); $image=$image['image']; header("Content-type: image/jpeg"); echo $image; } else echo"error"; ?>
  3. carlitosfigue

    Problem with Login form.

    Hello guys, first post here. I have a web system which contains a login form programmed in 3 different languages HTML, PHP and JS. The problem is that it's not working, you can access without entering any data, you just press enter and it will work, I don't know why it is not validating any credentials. I was thinking about some query problems but I don't know. I am a newbie on this. I have read a lot but haven't found an answer. A friend helped me build the system but left that uncompleted and he's nowhere to be found. I was wondering if you could help me out with this. <form role="form" ng-submit="login(user,password)"> <div class="form-group"> <input type="user" class="form-control" ng-model='user' placeholder="Usuario"> </div> <div class="form-group"> <input type="password" class="form-control" ng-model='password' placeholder="Contraseña"> </div> <div class="alert alert-warning" id='alert' style="display:none">Revise la informacion...</div> <div class="alert alert-danger" style="display:none" id='alertErr'>Error Usuario o Contraseña Erronea intentelo de nuevo</div> <button type="submit" class="btn btn-primary">Ingresar</button> </form> <?php require_once 'database.php'; $db = new Database(); $body = json_decode(file_get_contents('php://input')); $user =$db->query("SELECT * FROM usuario WHERE usua_login = '".$body->user."' AND usua_pass = '".$body->password."'"); if($user == false){ http_response_code(404); } else{ http_response_code(200); echo json_encode($user); } ?> 'use strict'; /** * @ngdoc function * @name belkitaerpApp.controller:MainCtrl * @description * # MainCtrl * Controller of the belkitaerpApp */ angular.module('belkitaerpApp') .controller('MainCtrl', function ($scope,$http,$location) { $scope.login = function(user,password){ console.log('Login...'); if(user =='' || password ==''){ $('#alert').show("slow"); setTimeout(function() { $('#alert').hide('slow'); }, 3000); } else{ $http.post('../serverSide/login.php',{user:user,password:password}).success(function(data){ console.log('OK!'); $location.path('/products'); }).error(function(data){ $('#alertErr').show("slow"); setTimeout(function() { $('#alertErr').hide('slow'); }, 3000); }); } } });
  4. Hi all ! Is it good practice to a have a separate database for a different group of people, say people in a different country. For e.g. If I run a franchisee business model in different countries, would it be a good thing to have a separate database based on the country? I was thinking of naming the database prepended by country name like USA_mydatabase, Canada_mydatabase and so on. Would that be ok ? What is the best practice in such cases. Thanks all !
  5. Hello guys. I got a problem that whenever you register on my page, the registration is successful, no errors, successful redirection to login page, but the registration does not write the information into database and I have no idea why... I'm sure that i'm connecting correctly, to the correct table, with the correct commands, but it kinda does not work... BTW (This registration and login and all worked a few weeks ago, but I got an sudden internal server error, so I had to delete and reupload all files, and I had to change database. I changed the database, created the same table with the same columns, also I overwrote ALL old database information to the new (password, dbname,name) and, so page works fine, but that registration does not...I'm including my code for registration and registration form) Registration process CODE: <?php include_once 'db_connect.php'; include_once 'psl-config.php'; $error_msg = ""; if (isset($_POST['username'], $_POST['email'], $_POST['p'])) { // Sanitize and validate the data passed in $username = filter_input(INPUT_POST, 'username', FILTER_SANITIZE_STRING); $email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_EMAIL); $email = filter_var($email, FILTER_VALIDATE_EMAIL); if (!filter_var($email, FILTER_VALIDATE_EMAIL)) { // Not a valid email $error_msg .= '<p class="error">The email address you entered is not valid</p>'; } $password = filter_input(INPUT_POST, 'p', FILTER_SANITIZE_STRING); // Username validity and password validity have been checked client side. // This should should be adequate as nobody gains any advantage from // breaking these rules. // $prep_stmt = "SELECT id FROM members WHERE email = ? LIMIT 1"; $stmt = $mysqli->prepare($prep_stmt); // check existing email if ($stmt) { $stmt->bind_param('s', $email); $stmt->execute(); $stmt->store_result(); if ($stmt->num_rows == 1) { $error_msg .= '<p class="error">A user with this email address already exists.</p>'; } $stmt->close(); } // check existing username $prep_stmt = "SELECT id FROM members WHERE username = ? LIMIT 1"; $stmt = $mysqli->prepare($prep_stmt); if ($stmt) { $stmt->bind_param('s', $username); $stmt->execute(); $stmt->store_result(); if ($stmt->num_rows == 1) { $error_msg .= '<p class="error">A user with this username already exists.</p>'; } $stmt->close(); } // TODO: // We'll also have to account for the situation where the user doesn't have // rights to do registration, by checking what type of user is attempting to // perform the operation. if (empty($error_msg)) { // Create salted password $passwordHash = password_hash($password, PASSWORD_BCRYPT); // Insert the new user into the database if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password) VALUES (?, ?, ?)")) { $insert_stmt->bind_param('sss', $username, $email, $passwordHash); // Execute the prepared query. if (! $insert_stmt->execute()) { header('Location: ../error.php?err=Registration failure: INSERT'); } } header('Location: ./continue.php'); } } and Registration form : <div class="register-form"> <center><h2>Registration</h2></center> <form action="<?php echo esc_url($_SERVER['PHP_SELF']); ?>" method="post" name="registration_form"> <center><p></p><input type='text' name='username' placeholder="Username" id='username' /><br></center> <center><p></p><input type="text" name="email" id="email" placeholder="Email" /><br></center> <center><p></p><input type="password" name="password" placeholder="Insert Password" id="password"/><br></center> <center><p></p><input type="password" name="confirmpwd" placeholder="Repeat Password" id="confirmpwd" /><br></center> <center><p></p><input type="submit" class="button" value="Register" onclick="return regformhash(this.form, this.form.username, this.form.email, this.form.password, this.form.confirmpwd);" /> </center> </form> </div> Anybody know where is problem?
  6. $results = mysql_query("$select", $link_id); while ($query_data = mysql_fetch_row($results)) { $link_id = mysqli_connect("$db_host","$db_user","$db_password","$db_database"); if ($mysqli->connect_errno) { echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; } $link_id = mysql_connect("$db_host","$db_user","$db_password"); if (mysql_select_db("$db_database", $link_id)); else { echo "connection failed."; } I am having to update MySQL to mysqli as my host is upgrading to PHP5.6 I have managed to convert and connect to the database converted to I cannot get the fetch results to work, can anyone help me convert the following code Many Thanks
  7. Hello guys, I do not know if anyone work or have worked with the "elFinder (file manager)" .. I incorporate "elFinder" to my platform .. I have the following question .. Have its connector: <?php error_reporting(0); // Set E_ALL for debuging include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderConnector.class.php'; include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinder.class.php'; include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeDriver.class.php'; include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeLocalFileSystem.class.php'; // Required for MySQL storage connector include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeMySQL.class.php'; // Required for FTP connector support include_once dirname(__FILE__).DIRECTORY_SEPARATOR.'elFinderVolumeFTP.class.php'; /** * Simple function to demonstrate how to control file access using "accessControl" callback. * This method will disable accessing files/folders starting from '.' (dot) * * @param string $attr attribute name (read|write|locked|hidden) * @param string $path file path relative to volume root directory started with directory separator * @return bool|null **/ function access($attr, $path, $data, $volume) { return strpos(basename($path), '.') === 0 // if file/folder begins with '.' (dot) ? !($attr == 'read' || $attr == 'write') // set read+write to false, other (locked+hidden) set to true : null; // else elFinder decide it itself } $opts = array( // 'debug' => true, 'roots' => array( array( 'driver' => 'LocalFileSystem', // driver for accessing file system (REQUIRED) 'path' => '/home/', // path to files (REQUIRED) 'URL' => dirname($_SERVER['PHP_SELF']) . '/home/', // URL to files (REQUIRED) 'accessControl' => 'access' // disable and hide dot starting files (OPTIONAL) ) ) ); // run elFinder $connector = new elFinderConnector(new elFinder($opts)); $connector->run(); I want him to make the call the "$screen" (folder will be created automatically by the user) of each user .. and not the "/home/" (as an example of put) .. the "$screen" comes from my platform, and I add the "elFinder" to it .. Imagine that each user creates "1/2/3 or 4, etc .." folders .. The user then enters the "elFinder" and see the folder created "1/2/3 or 4" (created on the left side of the folders ) and want only see your folders him. My Example.. It gives me error when I try to use these functions .. I use these functions for the "$screen" .. Of course, that I try connect only the folder .. but the result was 0 .. //za screen imeto function clean($string) { $string = str_replace(' ', '-', $string); // Replaces all spaces with hyphens. $string = preg_replace('/[^A-Za-z0-9\-]/', '', $string); // Removes special chars. return preg_replace('/-+/', '-', $string); // Replaces multiple hyphens with single one. } //end $srvname = mysql_real_escape_string(trim($_POST['servername'])); //ime na servera $port = (int)$_POST['ports']; //port na servera $ip = mysql_real_escape_string(trim($_POST['serverip'])); //serverip $screen = clean($srvname).$port.'_'.uniqid(); //screen imeto $dir = "/home/"; $opts = array( // 'debug' => true, 'roots' => array( array( 'driver' => 'LocalFileSystem', // driver for accessing file system (REQUIRED) 'path' => '$dir/$screen', // path to files (REQUIRED) 'URL' => dirname($_SERVER['PHP_SELF']) . '$dir/$screen', // URL to files (REQUIRED) 'accessControl' => 'access' // disable and hide dot starting files (OPTIONAL) ) ) ); This example of more folders.. $opts = array( 'roots' => array( array( 'driver' => 'LocalFileSystem', // driver for accessing file system (REQUIRED) 'path' => 'path/to/files/first_root', // path to files (REQUIRED) 'URL' => 'http://localhost/files/first_root/', // URL to files (REQUIRED) 'alias' => 'First home', // The name to replace your actual path name. (OPTIONAL) 'accessControl' => 'access' // disable and hide dot starting files (OPTIONAL) ), array( 'driver' => 'LocalFileSystem', 'path' => 'path/to/files/second_root', 'URL' => 'http://localhost/files/second_root/', 'alias' => 'Second home' ) ) ); But I need to be automatic, for each creation of user.. I use the variable "$screen" to create automatic folder of user.. but how to call the database of $screen ? the $screen exist only on my platform.. I don't know if you understand me.. Regards, Joob
  8. I have a simple user login where it inserts the user session into database like this (id, user_id, hash). It works fine. I can delete the session from the database when the user logs out. Now I realized something. When I close the browser window, it'll destroy the session(log out the user) as intended but it won't delete the session from the database. I was wondering if there is a way to do that?
  9. Here's what I am trying to do. Users Table user_id, sponsor_id, username, filled_positions, position_1, position_2, position_3, position_4, position_5 1 0 user 1 4 user 2 user 3 user 4 user 5 2 1 user 2 2 user 4 user 5 3 1 user 3 4 2 user 4 5 2 user 5 Above is a "Users" table. Here's what I am trying to do. Insert new users into the table. Say I already have the users table set up with 5 users. I want to add User 6. I want to loop through the users in the table and find the next empty position and update it with the new user id. In this scenario diagram above, the next empty position is Row 1 - position_5. The one after that is Row 2 - position_3 and then Row 2 - position_4...etc. It basically loops through rows and checks each position. So User 6 will be placed under Row 1 - position_5 and User 7 will be placed under Row 2 - position_3. How can one go on about doing that?
  10. I want to display 10 products from each category from database, I started to display the categories but how to make displaying the products from each one. Here is the code I make so far. I'm not sure am I doing it right with doble sql select , or it can be done only with one. <div class="inner shadow"> <?php $query = 'SELECT id, title_bg AS `title_cat` FROM categories'; $result = $this->db->query($query); ?> <?php foreach ($result->result() as $row): ?> <?php $title_cat = stripslashes($row->title_cat); ?> <div class="prod-sec"> <div class="prod-head"> <h2><?= $title_cat?></h2> <div class="clear"></div> </div> <?php $query1 = 'SELECT t1.id, t1.title_bg AS `Title`, t1.text_bg AS `Text`, t1.price, t1.discount, t1.category_id, t1.promo_page, t2.id AS FileID, t2.ext, FROM products t1 LEFT JOIN products_pictures t2 ON t1.id = t2.object_id LEFT JOIN categories t3 ON t3.id = t1.category_id WHERE t1.promo_page = 0 AND t1.is_active = 1 AND t3.title_bg = '$title_cat' ORDER BY RAND() LIMIT 10'; $result1 = $this->db->query($query1); ?> <div id="classeslist2"> <ul class="home_middle_products"> <?PHP foreach($result1->result() as $row1) { $f = 'files/products_first_page/' . $row1->id . '.jpg'; if(is_file(dirname(__FILE__) . '/../../' . $f)) { $img = site_url() . "files/products_first_page/".$row1->id.".jpg"; } else { $img = site_url() . "files/products/".$row1->id."/".$row1->FileID."_2.".$row1->ext; } $title = stripslashes($row1->Title); $text = character_limiter(strip_tags(stripslashes($row1->Text)),250); $title_url = getLinkTitle($title); $link = site_url()."products/product/".$row1->id."/{$title_url}"; ?> <li style="width: 185px; height: 270px; margin-left: 3px; margin-top: 10px;"> <div class="thumb"> <a href="<?=$link?>"><img src="<?=$img?>" alt="<?=$title?>" width="182" /></a> <div class="price"><?PHP echo product_price($row1, array('show_discount' => false,"show_old_price"=>false, 'show_label' => false, 'view' => 'no')); ?> </div> </div> <h2></h2> <h2><a href="<?=$link?>"><?=$title?></a></h2> <? $text = substr($text, 0, 100); ?> <? if (strlen($text) == 100) $text .= '...' ; ?> <p><?=$text?></p> </li> <?php } ?> <div class="clear"></div> </ul> </div> </div> <?php endforeach; ?> <div class="clear"></div> </div>
  11. I am building an e-commerce site and I am aiming to create a front end displaying my products with an option for customers to buy them, and have a content management system as a back end for an admin to edit product information. Currently I am storing information about my products on a mysql database. I access and display the product info using a while loop. Below is a simplified version of what I am doing without any html to style it. This code will go through the database and each iteration will go the to the next row and display the info about the next product. $query = mysql_query("SELECT * FROM truffleProducts"); while ($row = mysql_fetch_array($query)) { $id = $row['id']; $name = $row{'Name'}; $price = $row{'Price'}; $desc = $row{'Description'}; echo $id; echo $name; echo $price; echo $desc; } I have began to implement a 'buy' button so that customers are able to click on a button next to the product info and purchase it. However I have come across a problem which is where my program won't be able to tell which product you have selected as the number stored in the $id variable will just be the last product it has fetched from the database. I can't differentiate between all the product's buy buttons, so it will impossible to place an order for a customer with the current system I have. Can any one tell me how to get the id number of the specific product that a user has selected? I only started learning PHP a month or two ago so assume I know nothing
  12. hello dear folks good evening dear PHP-Freaks, well i am not sure if this question fits here - in miscellaneous. but i am pretty sure that many many calc and excel-experts are here in this great forum. what is aimed: want to create a calendar - with 365-chunks that are imported into a caleandar form see some ideas where i can import this i posted the example-calc spreassheet that i want to import. note it has got 365 lines - for 365 days of the next year 2015 what needs to be done ; whats needs to be achieved: i need to impoort the example into the calendar form. see some of the examples -.. if i am able to import the texts (of the collumn ) in a writer or word document each line of the calc needs to get on a sheed of the word or writer. i need to know how to arrange this export of calc into the writer document. if i need to explain it more thoroughiy - just let me know! greetings you matze see some examples of calendar templates Blank daily calendar - Templates Daily lesson planner (color, landscape) - Templates note: all i need to know is to be able to export from calc to any word or calc document... see the attached sample below... 2015_sample_version_.ods.zip
  13. Werezwolf

    Contacts Databases Examples

    So i was looking around and found that the contacts book (address book for Outlook) has the potential to store allot of data, and i mean more then what anyone would properly put in. Here is the list (100 Columns): Full_Name, Title, First_Name, Middle_Name, Surname, Initals, Suffix, Job_Title, Department, Organization, Email, Street_Address, City, State/Province/County, Postal_Code, Country, Telephone, Home_Tel., Fax, Other_Fax, Other_Email, Mobile, Pager, Info, Home_Post_Address, Home_Street_Address, Postal_Address, SMR_Address, Web_Site, Business_Street_2, Business_Street_3, Home_Street_2, Home_Street_3, Home_City, Home_State, Home_Postal_Code, Home_Country, Other_Street, Other_Street_2, Other_Street_3, Other_City, Other_State, Other_Postal_Code, Other_Country, Assistant's_Phone, Business_Fax, Business_Phone_2, Callback, Car_Phone, Company_Main_Phone, Home_Fax, Home_Phone_2, ISDN, Other_Phone, Primary_Phone, Radio_Phone, Telex, Account, Anniversary, Assistant's_Name_, Billing_Information, Birthday, Business_Address_PO_Box, Categories, Children, Company_Yomi, Directory_Server, E-mail_Type, E-mail_Display_Name, E-mail_2_Address, E-mail_2_Type, E-mail_Display_Name, E-mail_3_Address, E-mail_3_Type, E-mail_3_Display_Name, Gender, Government_ID_Number, Hobby, Home_Address_PO_Box, Internet_Free_Busy, Keywords, Language, Location, Manager's_Name, Mileage, Office_Location, Organizational_ID_Number, Other_Address_PO_Box, Prioity, Private, Profession, Referred_By, Sensitivity, Spouse, Surname_Yomi, User_1, User_2, User_3, User_4, Nickname I particularly like: Government_ID_Number (this could be handy) Language (I don't think that most of us would be Multi-Lingal) Mileage (Because i like the car and i want to by it at 300,000 KM) So what would you want to put in your database?
  14. Hi there guys, I've a little problem with inserting a file name into a database table. I can't see wich is the problem. The code is bellow and i think the problem is at INSERT INTO part. <?php $path = "./cv/"; $valid_formats = array("doc", "pdf"); if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST") { $name = $_FILES['photoimg']['name']; $size = $_FILES['photoimg']['size']; if(strlen($name)) { list($txt, $ext) = explode(".", $name); if(in_array($ext,$valid_formats)) { if($size<(20480*20480)) // Image size max 20 MB { $actual_image_name = time().$id.".".$ext; $tmp = $_FILES['photoimg']['tmp_name']; if(move_uploaded_file($tmp, $path.$actual_image_name)) { mysqli_query($mysqli,"INSERT INTO formular_client (client_cv = '$actual_image_name')"); } else echo "failed"; } else echo "Image file size max 20 MB"; } else echo "Invalid file format.."; } } ?> <input type="file" name="photoimg" id="photoimg" />
  15. Hi, i'm new to php and mysql so any helps will be greatly appreciated. I have the the code below but it looks like it's not working. I need the data id, firstname from table technictians and the total sales(add all sales in the database together) in the everyday_sale table. It's give me the total sale in the second query but not the id and firstname from the first query. If i take out the second query then the first query gave me the id and firstname just fine. It's seem like that two queries dont like each other some how. Thanks in advance and will consider give donation if got the problem solve. <HTML> <HEAD> <TITLE></TITLE> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script> <script src="onpage_submit.js"></script> <META name="description" content=""> <META name="keywords" content=""> <META name="generator" content="CuteHTML"> </HEAD> <BODY BGCOLOR="#FFFFFF" TEXT="#000000" LINK="#0000FF" VLINK="#800080"> <?Php $connection = new mysqli('localhost', 'root', 'Mylovev1@', 'usnailsandspa'); if ($connection->connect_errno > 0) { die ('Unable to connect to database [' . $connection->connect_error . ']'); } $sql = "SELECT * FROM technictians"; if (!$result = $connection->query($sql)) { die ('There was an error running query[' . $connection->error . ']'); } ?> <?Php $connection = new mysqli('localhost', 'root', 'Mylovev1@', 'usnailsandspa'); if ($connection->connect_errno > 0) { die ('Unable to connect to database [' . $connection->connect_error . ']'); } $query = "SELECT SUM(sale) FROM everyday_sale where technictian_id= '.$row['id'].'"; if (!$result = $connection->query($query)) { die ('There was an error running query[' . $connection->error . ']'); } ?> <?php $rows = $result->num_rows; // Find total rows returned by database if($rows > 0) { $cols = 3; // Define number of columns $counter = 1; // Counter used to identify if we need to start or end a row $nbsp = $cols - ($rows % $cols); // Calculate the number of blank columns echo '<table border="1" bgcolor="#E0F2F7" bordercolor="blue" width ="700" height ="700" align="center">'; while ($row = $result->fetch_array()) { if(($counter % $cols) == 1) { // Check if it's new row echo '<tr>'; } echo '<td><table align="center"><tr><td align="center"><font size="5" color="red"><b>'.$row['firstname'].'</b></font></td></tr><tr><td valign="top"><form action="salaries_add.php" method="post"> Sale:&nbsp&nbsp&nbsp&nbsp&nbsp <input type="text" id="num1" name="sale" style="width: 100px;" /></td></tr><tr><td valign="top"> Tip:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp <input type="text" id="num2" name="tip" style="width: 100px;" /></td></tr><tr><td valign="top"> Ticket#:<input type="text" id="ticket_number" name="ticket_number" style="width: 100px;" /></td></tr><tr><td valign="top" align="center"> <input type="hidden" name="technictian_id" value="' .$row['id']. '"> <input type="submit" value="Submit"></form></td></tr> <tr><td><hr></td></tr> <tr><td align="center"><font sie="5" color="maroon"><b>Earning Totals:</b></font></td></tr> <tr><td>'.$row['SUM(sale)'].'</td></tr> </table></td>'; if(($counter % $cols) == 0) { // If it's last column in each row then counter remainder will be zero echo '</tr>'; } $counter++; // Increase the counter } $result->free(); if($nbsp > 0) { // Add unused column in last row for ($i = 0; $i < $nbsp; $i++) { echo '<td> </td>'; } echo '</tr>'; } echo '</table>'; } ?> <script> </BODY> </HTML>
  16. Hi, I'm trying to let JavaScript check if a givin user exist in the database. It seems that the _check-user.php always returns 0, but if I fill in a name that doesn't exist, and echo out the result variable in JS, the echo will return 1. Is there someone who could help me? JavaScript part: function checkUser() { $.post("_check_user.php", { 'username': document.getElementById("username").value }, function(result) { if (result == 1) { document.getElementById("checkUser").className = "succes"; document.getElementById("checkUser").innerHTML = "Name is available"; }else{ document.getElementById("checkUser").className = "errormsg"; document.getElementById("checkUser").innerHTML = "Name is not available!"; } }); } _check-user.php: <?php include("config.php"); $result = mysql_query("SELECT login FROM users WHERE login='".clean_string($_POST["username"])."'"); if(mysql_num_rows($result)>0){ echo 0; }else{ echo 1; } ?>
  17. Hello everyone, i'm Osiris and i'm stuck with this PHP/HTML code, i'm fairly new at this so this might be a dumb question. i have a database, but i want to output all data into a table from the database using HTML codes. i had it so far that i could outpust the info in just plain text but it doesn't output in the HTML table. Can anyone see what i did wrong? This is my PHP code its in the body section <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'password'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if(! $conn ) { die('Could not connect: ' . mysql_error()); } $sql = 'SELECT `Name`, `Color`, `Type`, `Subtype`, `Power`, `Toughness`, `Manacost`, `Rarity`, `Expansion`, `Foil`, `Stock` FROM `mtgtabel`'; mysql_select_db('mtgstock'); $retval = mysql_query( $sql, $conn ); if(! $retval ) { die('Could not get data: ' . mysql_error()); } echo '<table class=mytable>'; echo "<tr><th >Name</th><th >Color</th><th >Type</th><th >Sub Type</th><th >Power</th><th >Toughness</th><th >Converted mana cost</th><th >Rarity</th><th >Expansion</th><th >Foil</th><th >Stock</th></tr>"; while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) { echo "<tr><td>"; echo $row['Name'] echo "</td><td>"; echo $row['Color']; echo "</td><td>"; echo $row['Type']; echo "</td><td>"; echo $row['Subtype']; echo "</td><td>"; echo $row['Power']; echo "</td><td>"; echo $row['Toughness']; echo "</td><td>"; echo $row['Manacost']; echo "</td><td>"; echo $row['Rarity']; echo "</td><td>"; echo $row['Expansion']; echo "</td><td>"; echo $row['Foil']; echo "</td><td>"; echo $row['Stock']; echo "</td></tr>"; } echo "</table>"; mysql_close($conn); ?> Thanks in advance!
  18. edward1111

    Connect to myphpadmin/ mysql

    I am trying to connect to my database on xammp, Im trying to find info how to connect and set up for people that sign in to my website they can regsiter for it.
  19. subhomoy

    Need advice in making databases

    Hello everyone, I'm going to create my school project and I really need your advice. My Project Summary I need to create a new website where the user can add new schools in my site. After creating, they can add students, add teachers, add principal and hell lot of things from their admin panel. And in the frontend they will get their own urls and can display their schools. Ex1: example.com/st_sebastians/ <------- School name. Ex2: example.com/st_augustin/ <------- School name. When the user will visit those url they will see all the contents of their respective schools. My Questions How can I create those database. For that I've thought of two methods. 1) I will create multiple database for every user/school dynamically using php and saves the record in it. (I searched on google that it is very prone to mysql injection.) 2) I will create a single database with everything related to their (school id) and stores in the single database. Example Student table id | school_id | name | roll | ---------------------------------------------- 1 | 1 | Subho | 123456 ---------------------------------------------- 2 | 5 | xyz | 236566 ---------------------------------------------- 3 | 45 | asfgf | 778219 . . . . . . . . Please note that the database are going to store whole lot of records and I don't want to it slow down. Any Help will be highly appreciated. Thank You In advance...
  20. Hi, I have been using Hostgator (shared) for years. Recently, last 4 weeks, almost every week for a day when ever I enter my websites, they give me an error like "Error: No Database Connection", then everytime I contact to Hostgator support and they fix it, but they don't tell me what is really going on with my database system. (They just say my websites are not hacked or something they are doing some update, that is the reason.) So my questions are; 1- What can be the problem that I always have an database connection error and my websites are down for some hours everyweek? (I m not a db expert, so can't figure it out) Is it same with other hostings too? (I have not used any others before) 2- Hostgator users: Do you face this kind of problems too? 3- I have seen the "Web Host List" topic, but i don't know if the list was updated, so I would like to know if you have any hosting suggestions that you don't face this kind of problems. Thank you in advance for your replies.
  21. I have a server script with which i have allowed the user to mark any item as favorite, but i also want another script through which the user can unfavorite the same item if they want and that item should be deleted from their favorite list. I have kept the same table for both favorite and unfavorite code, therefore i have used update query to update the details. For this purpose i have a code, but its not working, as i am new in the programming field would appreciate if someone could provide the correct codes <?php require_once('config.php'); $favorite = $_REQUEST['favorite']; $unfavorite = $_REQUEST['unfavorite']; $id=$_REQUEST['id']; $unfavoritedeal=mysql_query("SELECT * FROM favoritedeals where id='".$id."'"); //favoritedeals is the name of the table if($row=mysql_fetch_array($unfavoritedeal)) { $favorite=$row['favorite']; $unfavorite=$row['unfavorite']; } $myfavorite=(isset($_REQUEST['favorite'])?$_REQUEST['favorite']:$favorite); $myunfavorite=(isset($_REQUEST['unfavorite'])?$_REQUEST['unfavorite']:$unfavorite); $update = mysql_query("update favoritedeals set favorite = '".$myfavorite."', unfavorite = 1 where id = '".$id."'"); if(unfavorite="1" where id='".$id."') { "delete from favoritedeals WHERE id= '".$id."'"; } $posts[0]['message'] = 'favorite list updated'; $selectt = mysql_query("select * from favoritedeals where id = '".$id."'"); $results = mysql_fetch_assoc($selectt); $posts[0]['detail'] = $results; header('Content-type: application/json'); echo json_encode($posts); ?>
  22. I have been using the below method to access my database for some time now and just wanted to make sure im doing the correct thing. My application runs perfectly but im just trying to improve my programming skills. The classes are all loaded in with an autoloader so in theory using the classes below I could just call: echo User::getUsername(1); Like I say it works fine I would just love some feedback about what other people do and suggestions on something that might run better or looks cleaner. class Db { private static $db_read; private static $db_write; public static method read(){ if( self::$db_read == null ){ //create new database connection is it doesnt exist self::$db_read = new PDO(); } return self::$db_read; } public static method write() { if( self::$db_write == null ){ //create new database connection is it doesnt exist self::$db_write = new PDO(); } return self::$db_write; } class User { public static function getUsername($user_id){ $d = Db::read()->prepare('select * from user where id = ? '); $d->execute( array($user_id) ); $user = $d->fetch(); return $user['username']; } }
  23. Hi Guys, I'm new to OOP. I've mastered some basic syntax, but am wondering about an issue of design. To clarify in this example, I'm simply looking to pull Team data from a Database (I've not included my db class (PDO wrapper) although I know it is working. Although I have seen this sort of method below in a book (I believe and on the net) and it does work, I cannot help thinking coupling the DB class so tightly with the Team Class isn't a great approach. Can anyone give me feedback as to whether my approach is valid? While I can see the negative issue of tight coupling (i.e. changes to any database method would require multiple changes in Team (as team methods were added), I cannot really see an alternative way of doing this? I guess this is an issue of a design pattern or more advance OOP. Can anyone suggest an alternative way and/or more reading on making the coupling looser while still achieving my objectives? Should, for example, all DB functionality in getName be done during implementation? I initally thought pushing all db related functionality inside a Team method was wise and the best way then to add further methods, i.e. getResults method would be similar to getName but obviously with a different query and processing afterwards inside Team, but now I wonder if all should be in the implementation, or is there another approach? Thanks in advance. // create team class class Team { private $db; private $team_name; private $team_id; private $result; // in constructor lets pass DB object public function __construct($db) { $this->db = $db; } //function to get team name public function getName($team_id) { $this->team_id = $team_id; $this->db->query("SELECT team_name, nickname, founded FROM club WHERE team_id=:teamid"); $this->db->bind(':teamid', $this->team_id); if ($result = $this->db->single()); { return $result[]; } } }// end class // Implementation $team_id=1; //passed from user input // First Create Db Object with correct passed variables $database = new Database($server,$db_type); // Invoke DB connect method $database->connect(); // Now create Team object, pass it database object $team = new Team($database); // call Team method $team_display = $team->getName($team_id); // close db $database->closedb();
  24. 1. HTML FORM #for user to enter the data <html> <title>reg</title> <style type="text/css"> body { background-color: rgb(200,200,200); color: white; padding: 20px; font-family: Arial, Verdana, sans-serif;} h4 { background-color: DarkCyan; padding: inherit;} h3 { background-color: #ee3e80; padding: inherit;} p { background-color: white; color: rgb(100,100,90); padding: inherit;} </style> <form method="POST" action="login_back.php" enctype="multipart/form-data"></br> &nbsp<font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username"> </br></br> &nbsp<font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/> </br></br> &nbsp<font color="DarkCyan"> Last name:</font><input type="text" name="lastname"/> </br></br> &nbsp<font color="DarkCyan"> File: <input type="file" name="image"></font> </br></br> <input type="submit" value="Save and Proceed"> </form> </html> ---------- 2 STORING IN DATABASE #backend processing to store and retrieve data from db <?php #echo "<body style='background-color:rgb(200,200,200)'>"; session_start(); if( isset($_POST['username']) && isset($_FILES['image']) ) { $_SESSION['username']=$_POST['username']; $_SESSION['firstname']=$_POST['firstname']; $lastname=$_POST['lastname']; $file=$_FILES['image']['tmp_name']; $image_size=getimagesize($_FILES['image']['tmp_name']); if(!isset($file)) echo"please select an image"; else { $image_name=$_FILES['image']['name']; //grabing image name $image_size=getimagesize($_FILES['image']['tmp_name']); //getting image size } echo "</br>"; #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); #checking the available username $query = mysql_query("SELECT * FROM userdata WHERE username = '" . $_SESSION['username'] . "'" ); $ans=mysql_num_rows($query); if ($ans > 0) { echo "Username already in use please try another."; } else if($image_size==FALSE) { echo"That's not an image."; } else { #Insert data into mysql #1.Inserting user name & image into db $sql="INSERT INTO userdata(username, firstname, lastname, image)VALUES('" . $_SESSION['username'] . "', '" . $_SESSION['firstname'] . "', '$lastname','$image')"; $result1=mysql_query($sql); if($result1) { echo "</br>"; echo "Registration successful"; echo "</br>"; //displaying image $lastid=mysql_insert_id();//get the id of the last record echo "uploaded image is :"; echo "<img src='get.php?id=".$lastid."'>"; > this command has some mistake }#if insertion into db successful else { echo "Problem in database operation"; } }# else block of unique username n img }#end of isset ?> ---------- 3. GET.PHP #additional file that retrieve image from database <?php #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); if(isset($_REQUEST['id']) ) > this block of code is not runninng { $mid=(int)($_REQUEST['id']); $image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error()); $image=mysql_fetch_assoc($image); $image=$image['image']; header("Content-type: image/jpeg"); echo $image; } else { echo"error"; } ?>
  25. matt20687

    PHP/SQL Nightmare

    Hey, I have an issue where the below script doesnt appear to be working. It doesnt actually display anything on the page, it seems as though after the first PHP tag is just ignores the rest of the script. when looking at the page source all it shows is: <HTML> <HEAD> <TITLE>Logs</TITLE> </HEAD> <BODY> The actual code is below: <HTML> <HEAD> <TITLE>Logs</TITLE> </HEAD> <BODY> <?PHP $mysqlserver="localhost"; $mysqlusername="root"; $mysqlpassword="test"; $dbname = 'test'; $con=mysqli_connect($mysqlserver, $mysqlusername, $mysqlpassword, $dbname); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con, "SELECT * FROM forwardtable"); echo "<table border='1'> <tr> <th>Acc Num</th> <th>Send Date</th> <th>Send Time</th> </tr>"; while($rowContent = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $rowContent['Acc'] . "</td>"; echo "<td>" . $rowContent['Date'] . "</td>"; echo "<td>" . $rowContent['Time'] . "</td>"; echo "</tr>"; } echo "</table>" ?> </BODY> </HTML> If I remove the PHP from within the code the source code goes as it should and shows the </BODY> and </HTML> tags. Any ideas anyone? Thanks, Matt
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