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Found 35 results

  1. Hello everyone, I am a total newbie in PHP and so do here. I am trying to build a table consisting foreign key and primary key and it is showing error like "Parse error: syntax error, unexpected 'cities' (T_STRING) in line number 29". I have marked up my lines according to sublime text 3 editor. I am trying to create a database. 1 <?php 2 $servername = "localhost"; 3 $username = "username"; 4 $password = "password"; 5 $dbname = "newDB"; 7 // Create connection 8 $conn = new mysqli($servername, $username, $password, $dbname); 10 // Check connection 11 if ($conn->connect_error) { 12 die("Connection failed: " . $conn->connect_error);} 14 $firstname = $conn->real_escape_string($_REQUEST['firstname']); 15 $lastname = $conn->real_escape_string($_REQUEST['lastname']); 16 $address = $conn->real_escape_string($_REQUEST['address']); 17 $city = $conn->real_escape_string($_REQUEST['city']); 18 $country = $conn->real_escape_string($_REQUEST['country']); 19 $phone_number = $conn->real_escape_string($_REQUEST['phone_number']); 20 $email = $conn->real_escape_string($_REQUEST['email']); 22 // Attempt insert query execution 23 $sql1 = "INSERT INTO cities VALUES ('$city')"; 25 $sql2 = "INSERT INTO countries VALUES ('$country')"; 27 $sql3 = "INSERT INTO Contacts (firstname, lastname, address, city, country, phone, email) VALUES ('$firstname', '$lastname', '$address', $city, $country, '$phone_number','$email')"; 29 SELECT * FROM cities; SELECT * FROM countries; SELECT * FROM Contacts; if($conn->query($sql1) === true){ echo "City added successfully."; } else{ echo "ERROR: Could not able to execute " . $conn->error; } if($conn->query($sql2) === true){ echo "Çountry added successfully."; } else{ echo "ERROR: Could not able to execute " . $conn->error; } if($conn->query($sql3) === true){ echo "Çontact added successfully."; } else{ echo "ERROR: Could not able to execute " . $conn->error; } // Close connection $conn->close(); ?>
  2. 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 !
  3. 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.
  4. 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']; } }
  5. 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>
  6. Hi I have a question about managing data from forms and database, to be exact for safe input/output data from form input fields. Do i need some filters to remove code from input if user try to insert ? When i making database table i limiting chars and same in form. Here is a piece of code i use just for test and example : // connection to database $dbh = new PDO('mysql:host=localhost;dbname=test123', 'root', ''); $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // variables to insert into database $username = $_POST['username']; $password = $_POST['password']; $email = $_POST['email']; // query with prepare statements $stmt = $dbh->prepare("INSERT INTO members (username, password, email) VALUES (:username, :password, :email)"); $stmt->bindParam(":username", $username, PDO::PARAM_STR); $stmt->bindParam(":password", $password, PDO::PARAM_STR); $stmt->bindParam(":email", $email, PDO::PARAM_STR); $stmt->execute(); $lastId = $dbh->lastInsertId(); // checking if query is passed and data is inserted into dataabse if($lastId > 0) { echo 'Thank u for register.'; } else { echo 'Something went wrong, please try again.'; }
  7. 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 $image; } else echo"error"; ?>
  8. 1. HTML FORM #for user to enter the data <html> <title>reg</title> <style type="text/css"> body { background-color: rgb(200,200,200); color: white; padding: 20px; font-family: Arial, Verdana, sans-serif;} h4 { background-color: DarkCyan; padding: inherit;} h3 { background-color: #ee3e80; padding: inherit;} p { background-color: white; color: rgb(100,100,90); padding: inherit;} </style> <form method="POST" action="login_back.php" enctype="multipart/form-data"></br> &nbsp<font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username"> </br></br> &nbsp<font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/> </br></br> &nbsp<font color="DarkCyan"> Last name:</font><input type="text" name="lastname"/> </br></br> &nbsp<font color="DarkCyan"> File: <input type="file" name="image"></font> </br></br> <input type="submit" value="Save and Proceed"> </form> </html> ---------- 2 STORING IN DATABASE #backend processing to store and retrieve data from db <?php #echo "<body style='background-color:rgb(200,200,200)'>"; session_start(); if( isset($_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"; } ?>
  9. $results = mysql_query("$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
  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. Hi I’m trying to develop an online application that will allow ‘Candidates’ to book on courses that “Providers” have posted. The application is going to have 4 different types of users, please see attached ‘Figure 1.4 - JobSkilla User Permissions & Roles.jpg’ all of which have different information associated with them. On first signup of a provider they will complete a company profile and a user will be created as the owner associated with that “Provider” that user will then be able to invite their team members (Team associations) to share the same “Provider” data but with restricted access to areas such as billing / invoives. I’m going to be building the application using laravel 5.2, I still want to use laravels Auth but how can I allow multiple auths? I was thinking of having a table called “Auth” that could manage the logins with a forging key to the ‘candidates’, ‘providers’ table. What’s the best way of handling multiple logins that are associated with 1 “Provider”? I’ve attached a EER diagram with the current database design this may be incomplete. I wanted to ask for some guidance with the best way to program/db design the project. I would appreciate if anyone can give me a better solution or point me in the right direction. I don't mind paying for advice. JobSkilla EER Diagram.pdf
  12. 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" />
  13. 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?
  14. Hey guys, I wanted to share something with everyone. I hope this is helpful. This is basically a native solution to easily store PHP session data in a MySQL database. Session variables contain data that is saved for a specific user by associating the user with a unique identity. Typically, PHP would store session variables in a local file system on the server by default. While this may be acceptable to many people who are running small to moderate PHP applications, some larger applications that require load balancing would need to be run on multiple servers with a load balancer. In such cases, each server running PHP would need a way to ensure that sessions continue to work properly. One common way to achieve this is to override where PHP opens, reads, writes, and destroys the session variables so that it can perform these operations on a table inside of a MySQL database. When this is performed, the web application can gain advantages such as session management, session logging, and session interactions. I have provided my source code for your reference here: https://github.com/dominicklee/PHP-MySQL-Sessions Hope this helps someone out!
  15. 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...
  16. 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?
  17. HI, first time making a website... would really appreciate help with any errors in my code. Thanks! <div id="site_content"> <div class="sidebar"> <!-- form to select what books would like to be viewed, i.e. what category? --> <h3>Search Our Library!</h3> <div class="form_style"> <form action='titleSearch.php' method='GET'> <center> <h1>Search Any Book Title: </h1> <input type='text' size='90' name='search'> <input type='submit' name='submit' value='Search' > </center> </form> <form action='catSearch.php' method='GET'> <center> <h1>Search Book Categories:</h1> <?php // make an sql query to show all of the book titles $titleSQL = "select class, description from l_classification order by class"; // connect to the database require_once('dbconnect.php'); // execute the query $rsTitle = mysql_query($titleSQL); echo "<select name=\"class\">\n"; echo "<option value=\"all\">Show all books</option>"; // loop through all of the book titles in the record set // while while($row = mysql_fetch_array($rsTitle)) { $class = $row['class']; $description = $row['description']; echo "\t<option value=\"$class\">$class : $description</option>\n"; } // each record i'll display as an option in the form // end loop ?> <input type='submit' name='submit' value='Search' ></br></br></br></center> </form> </select> </div> </body> </html>
  18. Hi All, I facing a problem with Ms Access. When i set my datatype to "Short Text", the data display perfectly. just because of limitation of "Short Text", it can only handle 255. So i decided to change the datatype to "Long Text", but the text wont display our. FYI, im using for web base data and i am using PHP language. Thanks
  19. I have an html form that inserts a record into database. I also have a an html edit form that retrieves the data from the database. Normally everything works. But I just noticed that I have an issue if I use double quotes in the form field. It'll insert into the database fine. The problem arises when outputting the variable in the form field. If I echo it outside the form field, it'll show up fine with the double quotes. But inside the field, I only geta partial string. For eg. $variable_input = '5 3/4" Glitter Concealed Platform Pump'; The above will insert to the database. But if I retrieve it in the form field below, it'll return with one number only. <input type="text" name="title" value="5" /> Is there a way to fix the variable output of the wording that includes the double quotes?
  20. Say I have 2 tables. Table 1: Type(type_id, type_name) Table 2: Records(record_id, record_name, record_type) Under the "record_type" column, is it better to use type_id(eg. 10) or type_name(eg. Clothing & Watches)? If the answer is use the name, is it better to use the original name(Clothing & Watches) or a slug name(clothing-watches)?
  21. Hi all, I’m building a search engine to find courses online; I wanted to ask for some guidance with the best way to program/structure the project. I’m going to be building the application using laravel 5.2 I have 3 different types of users as follows Users Course Providers Advisors All of which have different information associated with them such as Users Table first_name last_name date_of_birth email password Course Providers – this needs to have multiple logins associated with the course provider company_name address_line_1 address_line_2 postcode tel Advisors company_name first_name last_name email password I still want to use laravels Auth but how can I allow multiple auths or am I best in using roles if so how would this work? What’s the best way of handling multiple logins that are associated with 1 company? I would appreciate if anyone can give me a better solution or point me in the right direction.
  22. Say I am uploading an image that gets resized. The resized image is a thumb. I have it's file path saved in the database and the image itself saved in a folder. Originally I was saving both to the database and folder. But now that I think about it, do I have to save the orginal image? Wouldn't I be saving up a lot of space if I only save the thumb image? What do you think?
  23. Not sure why I can't get this been trying to figure it out all evening :/ The included smconfig.php contains my database password. Any ideas, I know it's gotta be easy, I must be missing something. PDO connect.php: <?php session_start(); include 'smconfig.php'; $db_host = "127.0.0.1"; $db_username = "root"; $db_pass = "$dbpass"; $db_name = "golden_wand"; // PDO CONNECT $db = new PDO('mysql:host='.$db_host.';dbname='.$db_name,$db_username,$db_pass); $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); ?> http://www.golden-wand.com/members/tester.php <?php include "../Scripts/connect.php"; $email = "test@gmail.com"; $stmt1 = $db->prepare("SELECT id, activated, username, email, password, token FROM members WHERE email=:email LIMIT 1"); $stmt1->bindParam(':email',$email,PDO::PARAM_STR); $stmt1->execute(); $count = $stmt1->rowCount(); if($count > 0){ while($row = $stmt1->fetch(PDO::FETCH_ASSOC)){ $uid = $row['id']; $username = $row['username']; $email = $row['email']; $hash = $row['password']; $activated = $row['activated']; $token = $row['token']; } } echo "Before: <br>"; echo 'UID = '.$uid; echo '<br>Token = '.$token; echo '<br>Activated = '.$activated; echo '<br>Hash = '.$hash; $activated="1"; $token = "md5($hash)"; try{ $db->beginTransaction(); $updateSQL = $db->prepare("UPDATE members SET activated=':activated' WHERE id=':uid' LIMIT 1"); $updateSQL->bindParam(':activated',$activated,PDO::PARAM_STR); $updateSQL->bindParam(':uid',$uid,PDO::PARAM_INT); $updateSQL->execute(); $db->commit(); echo "<br><br><br><br>Update Successful<br><br><br><br>"; } catch(PDOException $e){ $db->rollback(); echo "<br><br><br><br>Update Failed<br><br><br><br>"; } $stmt2 = $db->prepare("SELECT id, activated, username, email, password, token FROM members WHERE email=:email LIMIT 1"); $stmt2->bindParam(':email',$email,PDO::PARAM_STR); $stmt2->execute(); $count = $stmt2->rowCount(); if($count > 0){ while($row = $stmt2->fetch(PDO::FETCH_ASSOC)){ $uid = $row['id']; $username = $row['username']; $email = $row['email']; $hash = $row['password']; $activated = $row['activated']; $token = $row['token']; } } echo "After: <br>"; echo 'UID = '.$uid; echo '<br>Token = '.$token; echo '<br>Activated = '.$activated; echo '<br>Hash = '.$hash; ?> I own this site: http://www.golden-wand.com/phpfreaks.txt
  24. I am finding that if I have "0"(zero) value in form select option, it won't select this option or submit data. If I change this value to any other number to text, it will work. Is there a way to fix this? I have to have an option where I am able to choose to submit "0" value to the database table. <option value="0" <?php if(empty($_POST['special'])) {} else { if($_POST['special'] == 0) { echo 'selected'; } } ?> >None</option>
  25. Hello I'm trying to check if 2 values exist in the database using php, Google didn't help... I need something like this : if($stmt = mysqli_prepare($db_connect,'QUERY TO CHECK IF USERNAME AND EMAIL EXIST')){ mysqli_stmt_bind_param($stmt, "ss", $user,$email); mysqli_stmt_execute($stmt); /* if username exist echo username exist if email exist echo email exist */ } else{/*error*/} thanks !
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