Using Javascript to run a php script to check username

Question

I've been trying to build a registeration form for a website I am building. I can do the basics but I want it to check the username availability without reloading the page.

JAVASCRIPT

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js" type="text/javascript" charset="utf-8"></script>
<script>
    $(document).ready(function()
    {
        $("#Username").focusout(function()
        {
            //Check if usernane if available
            var username = $("#Username").val();
            $.post("scripts/check_username.php", {username:  username}, function(data)
            {
                if(data == 'false')
                {
                    alert('Username not available');
                    $("#Username").setCustomValidity("This username is already taken!");
                }
                else
                {
                    alert('Username available');
                }
            });
            return false;
        });
    });
</script>

HTML

<form id="registerForm">
                <table>
                    <tr><td>Username</td><td><input id="Username" class='textInput' type='text' name='username' required></td></tr>

PHP SCRIPT

<?php
include 'open_connection.php';
$result = 'true';
$username = mysql_real_escape_string($_POST['username']);

$result = mysql_query("SELECT * FROM tblMembers WHERE Username='$username'");
while($row = mysql_fetch_array($result))
{
    $result = 'false';
}
echo $result;
?>

When I leave the textbox it says username available no matter what. I placed a username "test" in the database... no luck

Please help

And your question is? Also, mysql_* functions are deprecated; you should be using mysqli_*. — Danny Beckett
– Danny Beckett, Commented Sep 30, 2013 at 1:18

Hassan Khodadadeh · Accepted Answer · 2013-09-30 01:38:42Z

1

PHP:

$output = 'true';
$username = mysql_real_escape_string($_POST['username']);
$result = mysql_query("SELECT * FROM tblMembers WHERE Username='$username'");
while($row = mysql_fetch_array($result))
{
    $output = 'false';
}
echo $output;

Ans Script:

<script>
$(document).ready(function()
{
    $("#Username").focusout(function()
    {
        //Check if usernane if available
        $.post("scripts/check_username.php", {username:  $("#Username").val()}, function(data)
        {

            if(data  =='false')
            {
                    $("#Username").setCustomValidity("This username is already taken!");
                }
                else
                {
                    alert('Username available');
                }
            });
            return false;
        });
    });
</script>

By the way, don't use mysql_* function, they're deprecated. use Mysqli or PDO. Next thing is you forgot to put semi-colon that the end of your statements !

edited Sep 30, 2013 at 1:38

answered Sep 30, 2013 at 1:23

Hassan Khodadadeh

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6 Comments

Hassan Khodadadeh Over a year ago

That's your connection or your server!

Tom Hanson Over a year ago

It was an error with my side, it doesn't work either. sorry for the confusion

Tom Hanson Over a year ago

Im still getting username availble when I use "test"

Hassan Khodadadeh Over a year ago

Use two variables in php like I did otherwise you get what you get now!

Tom Hanson Over a year ago

I'm sorry I don't see where you used two variables... sorry to be a pain

|

Kyle Muir · Accepted Answer · 2013-09-30 01:25:31Z

0

You specify $("#Username").value do you mean to use $("#Username").val()?

$('#Username').value is probably returning you rubbish which will not exist in your DB.

answered Sep 30, 2013 at 1:25

Kyle Muir

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Comments

humbolight · Accepted Answer · 2013-09-30 02:20:31Z

0

Do you use Developer Tools or Firebug?

It would be easy to see where the issue lies if you examined the $.post to check_username.php:

(1) is the correct post request being sent? I think your data object should be {"username":username}

(2) is your script responding true? or something else? You only know it's not returning false by the way your if statement is structured.

answered Sep 30, 2013 at 2:20

humbolight

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1 Comment

humbolight Over a year ago

I'm not talking about your editor/IDE. I'm talking about looking at the code in action -- in the browser.

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