MySQL update query using AJAX and PHP

Question

Scenario: I'm trying to incorporate it so that when you click this button, it adds 1 to a value in the database.

I've read so many articles about AJAX today but no solution.

P.S. The query works fine directly from the command line.

This is what I've written so far but I think I'm completely missing something.

game.php

<script>
function logCountAdd(){
        var request = $.ajax({
            type: "GET",
            url: "logCountAdd.php"
        });
        request.done(function(msg ) {
            alert('Success');
            return;
        });
        request.fail(function(jqXHR, textStatus) {
            alert("Request failed: " + textStatus);
        });
};
</script>
</head>
<body>
<button type="button" onclick="logCountAdd()">Gather Resources</button>

logCountAdd.php

<?php
$connection = mysqli_connect("localhost","root","","users");

if (mysqli_connect_errno())
{
    echo 'NOT_OK';
}

mysqli_query($connection, "UPDATE uc_users 
        SET logCount = logCount + 1
        WHERE user_name='Gregory'";)

mysqli_close($connection);

echo 'OK';
?>

Problem: After I click the button, the value in the database does not change.

The Error Code: GET logCountAdd.php 500 (Internal Server Error)jquery-1.11.2.min.js:4 m.ajaxTransport.sendjquery-1.11.2.min.js:4 m.extend.ajaxgame.php:7 logCountAddgame.php:22 onclick

First question asked on here, sorry guys!

Sorry about that, updated it with a clear explanation of the problem! — Gregory Worrall
– Gregory Worrall, Commented Jan 2, 2015 at 18:32
Does your console give you error ? Can you open up the chrome console tab and see if there is any error in there ? — Maz
– Maz, Commented Jan 2, 2015 at 18:33

FuzzyTree · Accepted Answer · 2015-01-02 18:43:23Z

3

mysqli_query($connection, "UPDATE uc_users 
        SET logCount = logCount + 1
        WHERE user_name='Gregory'";)
                                  ^ misplaced semicolon

move it to after the close parenthesis. You may want to turn on error reporting to make debugging easier

error_reporting(E_ALL);
ini_set('display_errors','On');

edited Jan 2, 2015 at 18:43

answered Jan 2, 2015 at 18:34

FuzzyTree

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4 Comments

Gregory Worrall Over a year ago

Wow. Wow. HA! Thankyou. Jesus Christ.

Gregory Worrall Over a year ago

Honestly, such a rookie mistake. I've been tinkering with this file for hours, confused why it wouldn't work and completely overlooked that semicolon. Ugh.

FuzzyTree Over a year ago

@gregjw i suggest turning on errors (see php.net/error_reporting). you'll have an easier time debugging problems like this

Gregory Worrall Over a year ago

Thank you! I'll do just that.

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