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Hello, I'm finishing a content manager, but I have a little problem with a iframe. Let me explain: I have a url stored in a table in a database, which is the path to a page called "galeria.php", which is a dynamic gallery. This gallery is loaded on the site when the user clicks on a submenu button called "portfolio". The iframe loads the gallery when clicked the "portfolio" button. So far so good. Loads exactly when and how I want. The problem is that when we are not on the "portfolio" page, the iframe keeps the occupation of space, even having no content. This causes on the website, unnecessary scroll. In practice what I ask, is how to do that when the iframe has no content, it becomes "hidden" or not take up space on the page. Thank you

So I've followed this code, corrected about 12 error, talked to my hosting and I am so done. I have tried everything but the error won't go away. The code is pasted below. As it's really late any help would be appreciated that would make my day the next day... <?PHP //connect to server $connect = mysql_connect("localhost","cello10_import","brigite27"); //connection to database mysql_select_db("cello10_import"); //query the database $query = mysql_query("SELECT * FROM users WHERE cover_image = 'http://d1w7fb2mkkr3kw.cloudfront.net/assets/images/book/small/9781/2500/9781250038821.jpg' "); //fetch the results / convert into array WHILE($rows = mysql_fetch_array(query)): $cover_image = $rows['cover_image']; $title = $rows['title']; $author = $rows['author']; echo "$cover_image<br>$title<br>$author<br><br><br>"; endwhile; ?> The error experienced is: Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in /home/cello10/public_html/display.php on line 10 Thanks for your help

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

Working on a website directory that includes ratings for listed companies. Trying to obtain information from database that would put some companies in a higher category than others. Or at least, I believe that is the approach I should be taking. There are a few tricky issues. 1. Ratings are their own custom posts but are attached to company pages (parents) 2. Rating values I need to obtain are meta keys (1. total number of ratings & 2. rating average of EACH) 3. Ratings are based on four criteria which are summed and averaged for each rating submitted. As a result, values are rarely single digits such as 4 or 5 but include decimals rounded to 1 place (e.g. 4.8 or 3.5) Additional info: Ratings are based on 5-star system (out of 5) Higher category would be business that achieved at least 10 ratings of 4 or more Any help appreciated.

i want a situation where, when i upload something into my database, it creates a new table, everytime, i saw this $forum = "CREATE TABLE IF NOT EXISTS for_".$id." ( `id` int(11) NOT NULL AUTO_INCREMENT, `stud` int(11) DEFAULT NULL, `course` int(11) DEFAULT NULL, `message` varchar(1000) DEFAULT NULL, PRIMARY KEY (`id`) ) "; but don't know how to implement it into this query $sql = "INSERT into data(name,Groups,phone_number) values('$data[0]','$data[1]','$data[2]')";

I have created a system that uses an API to drop utility data into MSQL Server 2008. Once the data is in the db, a web app can query the data to create electric bills. Currently we are working with only one customer but are entertaining expansion and want to approach the growth from the best possible angle. Considering that all data is the same type of data but comes from different customers and sources, would it be best to maintain 1 database with all of the utility meter data streaming to it or would it be best to have a new data base with each customer? Also, assuming we keep it all in the same database, would it be best (assuming we are secure) to have all the individual customers data go to the same tables or have separate tables for separate customers. I'm pretty certain I know the answer but I'm a novice so that means nothing. Thank you for any help with this issue.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Import a CSV File with PHP & MySQL</title> </head> <body> <form action="" method="post" enctype="multipart/form-data" name="csv" id="csv"> Choose your file: <br /> <input name="csv" type="file" id="csv" /> <input type="submit" name="Submit" value="Submit" /> </form> </body> but i keep having this error Notice: Undefined index: csv in C:\wamp\www\youngsoul\upload\file.php on line 7

Hi Guys I need some help with DB design and implementation of a feature my next website will have. OK so I'm building a system that will warn the current user that the current client needs a certain test before they can have their next treatment. Not all treatments required tests. So far I have the following tables user treatment treatmentTest treatmentUser iduser idtreatment idtreatmentTest idtreatment idtreatmentTest treatmentTestExpiry iduser dateOfTreatment (stored in yyyy-mm-dd) (the treatmentTestExpiry stores how long until the treatment will expire e.g +1 month) So every time a user comes in for a treatment we log the treatment in the treatmentuser table. This works fine and I can see what treatments the user has had. So certain treatments required a test to be done before the user can have that actual treatment. What I want is when you view the users profile page it says if they have not had a particular test or a previous test has expired and they will need another. I have no idea how to go about this. Im just wondering if someone could help talk me through it?

Hi guys, I have a problem with my code. It connects with the database but it doesn't execute anything. I am trying to update a value in a table but I really don't know what I am doing wrong. I am trying to change the price based on the meta_value GF-1370 Thank you very much. Here is Table (Taken from myphpadmin) meta_id post_id meta_key meta_value 18538 4356 _sku GF-1370 18541 4356 _price 2.343 and here is my code: <?php $link = mysql_connect('server', 'username', 'password'); if (!$link) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_select_db(wordpress_i12dhflgoa); $sql="update `wp_pzvy_postmeta` set `_price` = 5 WHERE `_sku` = 'GF-1370'"; ?>

Given the below 3 database tables I am trying to construct a SQL query that will give me the following result: customer_favourites.cust_id customer_favourites.prod_id OR product.id product.code product.product_name product.hidden product_ section.section_id (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.catpage (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.relative_order (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) I currently have.... SELECT customer_favourites.cust_id, customer_favourites.prod_id, product.code, product.product_name, product.hidden, product_section.section_id, product_section.relative_order, product_section.catpage FROM customer _favourites INNER JOIN product ON customer_favourites.prod_id = product.id INNER JOIN product_section ON product_section.product_code = product.code WHERE `cust_id` = '17' AND `hidden` = '0' GROUP BY `code` ORDER BY `section_id` ASC, `relative_order` ASC, `catpage` ASC LIMIT 0,30 This gives me what I want but only sometimes, at other times it randomly selects any row from the product_section table. I was hoping that by having the row I want as the last row (most recent added) in the product_section table then it would select that row by default but it is not consistent. Somehow, I need to be able to specify which row to return in the product_section table, it needs to be the row with the lowest section_id value or it should by the last row (most recent). Pulling my hair out so any help is gratefully received. customer_favourites id cust_id prod_id 70 4 469 product id code product_name hidden 469 ABC123 My Product 0 product_section id section_id catpage product_code relative_order recommended 44105 19 232 ABC123 260 1 44106 3 125 ABC123 87 1 44107 2 98 ABC123 128 1 44108 1 156 ABC123 58 0

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!

<?php if(isset($_POST['submit'])){ $uname = $_POST['username']; $pword = $_POST['password']; /*** mysql hostname ***/ $hostname = 'localhost'; /*** mysql username ***/ $username = 'root'; /*** mysql password ***/ $password = 'anty90'; try { $link = new PDO("mysql:host=$hostname;dbname=gambling", $username, $password); /*** echo a message saying we have connected ***/ echo 'Connected to database<br />'; /*** INSERT data ***/ $stmt = $link->prepare("INSERT INTO gamb(username, password) VALUES (?, ?)"); try{ $stmt->execute(array("$uname", "$pword")); } catch(PDOException $e){ echo "Exception caught: $e"; } /*** echo the number of affected rows ***/ //echo $count; /*** close the database connection ***/ $link = null; } catch(PDOException $e) { echo $e->getMessage(); } } ?> <html> <form action='home.php' method='post'> <input type="text" name="username" > <input type="password" name="password" > <input type="submit" name="submit" value="submit"> </form> </html> I'm new to databse programming so I was just wondering if this was vulnerable to sql injection or not.

(I'm putting this in PHP since it's not a question specific to MySQL or other DB stuff.) I have a page that uses the GET id to find a product. GET variables are sanitized, and the SQL string is escaped even though it's expecting a number only. So the code seems safe to me. I'm getting some error_log results that appear to be hack attempts: SELECT p.*, t.id as blah FROM some_table p left outer join some_other_table t on p.id = t.product_id WHERE p.id = 139\' and benchmark(20000000,sha1(1))-- Should I be worried about something like this? Anything more (or less) that I should be doing?

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

hello I wanna replicate one table in a database test(in localhost-master) to another database table(slave) on remote server(on the internet) in godaddy.com. I was really tried my best to make it work. but it doesn't seems to be working. I red the MySQL documentation as well. database name- test master table - attendance -on the localhost(127.0.0.1) slave table - attendance - on the internet(godaddy 184.x.x.x) I want to copy data from my attendance table on the localhost to slave table on the internet. Please help me to configure these settings. Thank you

mysql.php <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ckeditor",$con); ?> --------------------------------------------------------- add.php <?php include("mysql.php"); if(isset($_POST["button2"])) { $sql="INSERT INTO cktext (section) VALUES ('$_POST[select2]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } ?> ----------------------------------------------------------------- home.php <form id="form1" name="form1" method="post" action="add.php"> <tr> <td>Section:</td> <td><select name="select2" id="select2"> <option selected="selected" value="MALE">Male</option> <option selected="" value="FEMAIL">FEMAIL</option> </select></td> </tr> <input type="submit" name="button2" id="button2" value="Upload" /> </form> In DATABASE :- cktext table attribute "section" is varchar type. BUT IT RETURN ME BLANK OUTPUT.

Hello, my name is Casper. I'm createing a site where you can upload articles. But for now I onley can show 1 article on the pages. Does anyone knows how to show more articles? My script what posts 1 article: <?php $username = ""; $db = ""; $hostname = ""; $password = ""; @mysql_connect($hostname, $username, '') or die("connectie mislukt"); @mysql_select_db("$db") or die ("could not connect dattabase"); $sql = mysql_query('SELECT * FROM home ORDER BY Date') or die ('Database Error()'); $id = 'ID'; $date = 'Date'; $title = 'Title'; $article = 'Article'; $rows = mysql_fetch_assoc($sql) or die ('Fetch Data Error()'); echo $rows[$title], '-', $rows[$date]; echo '</br>'; echo '</br>'; echo $rows[$article]; ?> and here's a ss of my DB table: http://oi59.tinypic.com/2monptv.jpg every article has to be in this order: title - date article thank you in advance, Casper Hollemans.

Hello everybody, This is my first post here and I am beginner in PHP world in terms of writing the code to serve my own purposes. Well I am building a web app and basically it's a calendar which pulls information from .js file. Now I have thinking for the past couple of days how can I accomplish that each user that registers on the site manipulates its own .js file because information from .js file will be shown on calendar. Let me tell you how it's currently set up: 1. JavaScript file with particular static name is called under the script that is placed on index.php and the data is displayed on the page itself. So I would love to have is set it up like this: 1. Index page contains login form - Each registered/logged in user will have its own session 2. User registers and based on username/email new .js file is created out of a blank template and it is named based on user's username 3. user is then redirected to the calendar index which contains javascript that cals out that appropriate .js file based on the what user is logged in and displays data to the calendar I am not sure if that is doable with PHP or not but that's my thinking how it can be done if it's doable. I am open for any kind of suggestions how all this can be put together and if you do have better ideas I would love to hear from you.

Hi, I am having troubles trying to add View/Edit/Del options to my web system modules. I am not the creator of the web system so it's complicated for me, also, I am a newbie on this. I want to be able to view/edit/del from admin panel every product or client stored in my db. Here's the Clients module code: client.html <div ng-include="'views/menu.html'"> </div> <button type="button" class="btn btn-primary" ng-model = 'newProduct' ng-click = 'newProductButton()'>{{buttonName}}</button> <hr> <div class="table-responsive" ng-show ='!newProduct'> <table class="table table-striped table-hover"> <tr> <td>Nº</td> <td>Cédula</td> <td>Nombre</td> <td>Apellido</td> <td>Dirección</td> <td>Email</td> <td>Télefono</td> </tr> <tr ng-repeat= 'client in clients track by $index'> <td>{{$index + 1}}</td> <td>{{client.codigo}}</td> <td>{{client.nombre}}</td> <td>{{client.apellido}}</td> <td>{{client.direccion}}</td> <td>{{client.email}}</td> <td>{{client.telefono}}</td> </tr> </table> </div> <div ng-show = 'newProduct'> <form role="form" ng-submit ="updateClient (id,name,lastName,address,phoneNumber,email)"> <label>Cédula</label> <div class="form-group"> <input class="form-control" placeholder="Cédula" ng-model='id' id="id"> </div> <label>Nombre</label> <div class="form-group"> <input class="form-control" placeholder="Nombre" ng-model='name' id="sname"> </div> <label>Apellido</label> <div class="form-group"> <input class="form-control" placeholder="Apellido" ng-model='lastName' id="lastName"> </div> <label>Dirección</label> <div class="form-group"> <input class="form-control" placeholder="Dirección" ng-model='address' id="address"> </div> <label>Teléfono</label> <div class="form-group"> <input type='tel' class="form-control" placeholder="Teléfono" ng-model='phoneNumber' id="phoneNumber"> </div> <label>Email</label> <div class="form-group"> <input type='email'class="form-control" placeholder="Email" ng-model='email' id="email"> </div> <div class="alert alert-success" id='alertSuccess' style="display:none">Ingresado Satisfactoriamente...</div> <div class="alert alert-danger" id='alertDanger' style="display:none">Ese cliente ya fue agregado</div> <button type="submit" class="btn btn-primary">Agregar Cliente</button> </form> </div> client.js 'use strict'; /** * @ngdoc function * @name belkitaerpApp.controller:ClientCtrl * @description * # ClientCtrl * Controller of the belkitaerpApp */ angular.module('belkitaerpApp') .controller('ClientCtrl', function ($scope,$http) { $scope.newProduct = false; if($scope.newProduct){ $scope.buttonName = 'Ver Tabla'; } else{ $scope.buttonName = 'Ingresar Cliente'; } $http.get('../serverSide/clients.php').success(function(clients){ console.log(clients); $scope.clients = clients.Clients; }) $scope.newProductButton = function(){ $scope.newProduct = !$scope.newProduct; if($scope.newProduct){ $scope.buttonName = 'Ver Tabla'; } else{ $scope.buttonName = 'Ingresar Cliente'; } } $scope.updateClient = function(id,name,lastName,address,phoneNumber,email){ $http.post('../serverSide/updateClient.php',{id:id,name:name,lastName:lastName,address:address,phoneNumber:phoneNumber,email:email}).success(function(data){ console.log(data); $('#alertSuccess').show("slow"); setTimeout(function() { $('#alertSuccess').hide('slow'); }, 3000); }).error(function(data){ console.log(data); $('<div id="alertDanger"></div>').show("slow"); setTimeout(function() { $('<div id="alertDanger"></div>').hide('slow'); }, 3000); }) } }); client.php <?php require_once 'database.php'; $db = new Database(); $clients = $db->queryAll('SELECT clie_id as id,clie_cod as codigo, clie_ape as apellido, clie_nom as nombre, clie_dir as direccion, clie_ema as email, clie_tel as telefono FROM cliente','Clients'); echo json_encode($clients); ?> I have found a php code on the web and I was wondering if it was possible to merge it with my existing one. I tested the code with my db and it's working but I do not know how to merge it with my existing code. Here's the code: <?php if (isset($_POST['submit'])) { include 'db.php'; $clie_cod=$_POST['clie_cod'] ; $clie_ape= $_POST['clie_ape'] ; $clie_nom=$_POST['clie_nom'] ; $clie_dir=$_POST['clie_dir'] ; mysql_query("INSERT INTO `cliente`(Código,Apellido,Nombre,Dirección) VALUES ('$clie_cod','$clie_ape','$clie_nom','$clie_dir')"); } ?> </form> <table border="1"> <?php include("db.php"); $result=mysql_query("SELECT * FROM cliente"); while($test = mysql_fetch_array($result)) { $id = $test['clie_id']; echo "<tr align='center'>"; echo"<td><font color='black'>" .$test['clie_id']."</font></td>"; echo"<td><font color='black'>" .$test['clie_cod']."</font></td>"; echo"<td><font color='black'>". $test['clie_ape']. "</font></td>"; echo"<td><font color='black'>". $test['clie_nom']. "</font></td>"; echo"<td><font color='black'>". $test['clie_dir']. "</font></td>"; echo"<td> <a href ='view.php?clie_id=$id'>Edit</a>"; echo"<td> <a href ='del.php?clie_id=$id'><center>Delete</center></a>"; echo "</tr>"; } mysql_close($conn); ?> </table> </body> </html> Any help would be highly appreciated. Thank you very much.

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); }); } } });

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); ?>

function get_checked($variable, $status){ if($variable == $status){ return 'checked="checked"'; }else{ return ''; } } <input type=\"checkbox\" name=\"emailNewsletter\" value=\"Yes\" ".get_checked($emailNewsletter, 'Yes')."> <input type=\"checkbox\" name=\"emailInbox\" value=\"Yes\" ".get_checked($emailInbox, 'Yes')."> $emailNewsletter = isset($_POST['emailNewsletter']) ? filter_input(INPUT_POST, 'emailNewsletter') : 'No'; $emailInbox = isset($_POST['emailInbox']) ? filter_input(INPUT_POST, 'emailInbox') : 'No'; If no submit button clicked display mysql information $userInfo['emailInbox'] instead of 'NO' . How to do that?

helllo dear php-commmunity new to Ruby - i need some advices - i plan to do some requests in osm-files. (openstreetmap) Question - how can i store the results on a Database - eg mysql or - (if you prefer postgresql) - note: my favorite db - at least at the moment is mysql here the code require 'open-uri' require "net/http" require 'rexml/document' def query_overpass(object_type, left,bottom,right,top, key, value) base_url = "http://www.overpass-api.de/api/xapi?" query_string = "#{object_type}[bbox=#{left},#{bottom},#{right},#{top}][#{key}=#{value}]" url = "#{base_url}#{URI.encode(query_string)}" resp = Net::HTTP.get_response(URI.parse(url)) data = resp.body return data end overpass_result = REXML::Document.new(query_overpass("node", 7.1,51.2,7.2,51.3,"amenity","restaurant|pub|ice_cream|food_court|fast_food|cafe|biergarten|bar|bakery|steak|pasta|pizza|sushi|asia|nightclub")) overpass_result.elements.each('osm/node') {|x| if !x.elements["tag[@k='name']"].nil? print x.elements["tag[@k='name']"].attributes["v"] end print " | " if !x.elements["tag[@k='addr:postcode']"].nil? print x.elements["tag[@k='addr:postcode']"].attributes["v"] print ", " end if !x.elements["tag[@k='addr:city']"].nil? print x.elements["tag[@k='addr:city']"].attributes["v"] print ", " end if !x.elements["tag[@k='addr:street']"].nil? print x.elements["tag[@k='addr:street']"].attributes["v"] print ", " end if !x.elements["tag[@k='addr:housenumber']"].nil? print x.elements["tag[@k='addr:housenumber']"].attributes["v"] end print " | " print x.attributes["lat"] print " | " print x.attributes["lon"] print " | " if !x.elements["tag[@k='website']"].nil? print x.elements["tag[@k='website']"].attributes["v"] end print " | " if !x.elements["tag[@k='amenity']"].nil? print x.elements["tag[@k='amenity']"].attributes["v"] print " | " end puts } look forward to hear from you again - i would love to store it on a mysql - database - if possible. If you would prefer postgresql - then i would takte this one.... ;-) look forward to hear from you again - i would love to store it on a mysql - database - if possible. If you would prefer postgresql - then i would takte this one.... ;-) well - i guess that the answer to this will be the same no matter what language we are using. If the db is a sql database we need to design the database schema and create the tables in the database. The first step in accessing a db in our code is to get a connection to it. If ruby is our choice of language, a search for "ruby sql connector" will give us lots of options to read about. Well - we also can do it in PHP. What do you think!? Next, based on the schema we have designed, we need to create queries suitable for storing the data. We will likely need to consider our transactional model. Again, searching "ruby sql transactional model" will give us plenty of food for thought. Finally, we may want or need to close the connection to the database.

I built a CMS system using CKEditor and KCFinder that store information od a databse via textarea/php. So far so good! The issue comes to when I want to store and display images that link to themselves. The way I am storing images is exactly the same: There is a textarea where I insert an image via KCFinder/CKEditor. The image is uploaded to the server and the path stored at the database. Later I try to pick up that path from the database to display the image and because I want the image to link to itself, I try to use the same method to insert the url on the link. Problem? The link is missing and the images are not displaying. Can anyone point me the error and suggest any solution? I would be so thankful! CODE: try { $DBH = new PDO('mysql:host=localhost;dbname=yourdb;charset=utf8', 'user', 'password'); $DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $STH = $DBH->prepare('SELECT * FROM php_maskiner ORDER BY timestamp DESC'); $STH->execute(); $STH->setFetchMode(PDO::FETCH_OBJ); while($row = $STH->fetch()) { $title = $row->title; $entry = $row->entry; $images = $row->images; $img_url = $row->images; $img_pack = '<div class="mask3 span3"> <a rel="prettyPhoto" href="'.$img_url.'"><img src="'.$images.'"></a> </div>'; } $DBH = null; } catch (PDOException $e) { echo '<div class="alert alert-standard fade in"> <a class="close" data-dismiss="alert" href="#">×</a> <strong>Can\'t read the database!</strong> </div><br />'.$e; } <?php echo '<article class="span12 post"> '.$img_pack.' <div class="inside"> <div class="span8 entry-content"> <div class="span12"> <h2>'.$title.'</h2> <p>'.$entry.'</p> </div> </div> </div> </article>'; ?> Thanks in advance!

Hey there I have to make a php driven website that uses the CRUD operations to modify a simple MySQL database anyone willing to help me with this in detail? PM me if possible please thanks