Updating a database with jQuery using AJAX

Question

I'm having trouble getting this AJAX code to update my database. The code is a image that onClick will run the command to update the database

HTML:

<a>
    <img class       = "heart" 
         src         = "images/heart.png" 
         onClick     = "favUpdate(0,1)" 
         onMouseover = "this.src='images/heart_mo.png'"
         onMouseout  = "this.src='images/heart.png'"/>
</a>

Javascript code:

function favUpdate(fav_up, id_up) {
        $.ajax({
            type: 'post',
            url: 'includes/fav_update.php',
            data: {favorite: fav_up, id: id_up},
            success: function(output) {
              alert('success, server says '
                            + output
                            + 'Variables passed are '+fav_up+' '+id_up);
                }, 
                    error: function() {
              alert('something went wrong, Favorite update failed');
            }
            });
}

PHP code:

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_POST['favorite']);
    $fav_id     = mysql_real_escape_string($_POST['id']);
    $query      = "UPDATE projects SET favorite = $fav_update WHERE id = $fav_id";
    mysql_query($query, $main); 
?>

main.php

<?php
$hostname_main = "localhost";
$database_main = "test";
$username_main = "root";
$password_main = "";
$main = mysql_pconnect($hostname_main, $username_main, $password_main) or trigger_error(mysql_error(),E_USER_ERROR); 
?>

Does anyone know why it's not updating the database and why the "option" isn't getting data for the variable?

check that query run successfully and show us if you getting any error — NullPoiиteя, Apr 01 '13 at 04:41
Try `"UPDATE projects SET favorite = '".$fav_update."' WHERE id = '".$fav_id."'";` — Dipesh Parmar, Apr 01 '13 at 04:41
also you are using obsolesce `mysql_*` api its deprecated and will cause E_DEPRECATED error in php >=5.5 check http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php/14110189#14110189 — NullPoiиteя, Apr 01 '13 at 04:43
the server i'm using has php version 5.3.19, so i dont believe thats the issue, i tried to run the query in PHPMyAdmin and worked perfectly fine — raul5660, Apr 01 '13 at 04:55

score 1 · Answer 1 · answered Apr 01 '13 at 04:46

Try this

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_POST['favorite']);
    $fav_id = mysql_real_escape_string($_POST['id']);
    $query = "UPDATE projects SET favorite = '".$fav_update."' WHERE id = '".$fav_id."'";
    mysql_query($query, $main); 
?>

Yogesh Suthar · Answer 2 · 2013-04-01T04:57:42.040

0

you have to put single quote around $fav_update if its datatype is string VARCHAR,TEXT

$query = "UPDATE projects SET favorite = '$fav_update' WHERE id = $fav_id";
                                         ^           ^

Remove $main from here and try

mysql_query($query);

edited Apr 01 '13 at 04:57

answered Apr 01 '13 at 04:41

Yogesh Suthar

30,424
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on the alert(output) its echoing the query and the query is right and works on PHPMyAdmin – raul5660 Apr 01 '13 at 05:30
@raul5660 r us sure you have made the connection properly with database? – Yogesh Suthar Apr 01 '13 at 05:32
yes because i've been using the same connection file for multiple scripts before this, thats why i'm stumped – raul5660 Apr 01 '13 at 05:34

score 0 · Answer 3 · answered Apr 01 '13 at 04:57

0

Please try to debug your PHP (server side code first) :-

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_REQUEST['favorite']);
    $fav_id = mysql_real_escape_string($_REQUEST['id']);
    $query = "UPDATE projects SET favorite = $fav_update WHERE id = $fav_id";
    echo $query;
     mysql_query($query, $main); 
?>

Use $_REQUEST instead of $_POST, and call this api directly from the browser, by creating its url like http://localhost/filename.php?favorite=somevalue1&id=somevalue2

And check whether you got the insert in DB or not, and check the query by printing it too.

And after checking the API, please change the $_REQUEST, back to $_POST

answered Apr 01 '13 at 04:57

Nishant

3,614
1
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i'm getting the right query but database still isnt being updated? – raul5660 Apr 01 '13 at 05:14
please cehck your main.php once, or post that file over here – Nishant Apr 01 '13 at 05:39
i've checked it but the funny thing is that it works cause i actually use that connection file frequently thru out the site – raul5660 Apr 01 '13 at 05:43

score 0 · Answer 4 · answered Apr 01 '13 at 16:39

Changed php script to:

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_POST['favorite']);
    $fav_id = mysql_real_escape_string($_POST['id']);
    $updateSQL = sprintf("UPDATE projects 
                          SET favorite=%s  
                          WHERE id=%s",
                          $fav_update,
                          $fav_id);
    mysql_select_db($database_main, $main);
    $Result1 = mysql_query($updateSQL, $main) or die(mysql_error());


?>

hope that helps if anyone runs into the same issue

Updating a database with jQuery using AJAX

4 Answers4