CODEWITHSUNDEEP
phphtmljquery
raul5660
raul5660

Reputation: 11

Updating a database with jQuery using AJAX

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?

Upvotes: 0

Views: 515

Answers (4)

raul5660
raul5660

Reputation: 11

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

Upvotes: 0

Nishant
Nishant

Reputation: 3704

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

Upvotes: 0

Yogesh Suthar
Yogesh Suthar

Reputation: 30488

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);

Upvotes: 0

Kautil
Kautil

Reputation: 1331

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); 
?>

Upvotes: 1

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