It turns out James it was that the file that ‘action’ was pointing to, it used a session ‘lastUrl’ which in some instances was ‘checkUsername.php’! When the form failed any field validation it redirected to this – hence the blank screen.

Nice one, thanks, great script Sachin!

Check the ‘action=’ bit in your form declaration. Sounds to me (just a guess) that you have something like

You should be pointing it somewhere else, like maybe action=”join.php”

-James.

Great and simple script, thanks!

I’ve wrapped this up in a form with a submit button. Without clicking the availability button and clicking submit everything works fine. If I click the availability button, the script works, but if I follow this with my submit button a get a blank screen with {“code”:-1,”result”:”Invalid username”}.

It’s as if it’s calling my onclick event, any ideas?

Cheers.

Eg. What exactly does the eval part do and where can I go to find alternatives?

This is fairly simple to do. You can remove the button and instead associate the JavaScript function onblur event of input text, instead of onclick for the button.

thx for the code it really helps.
however in your code, we gotta click on the “check availability” button first to validate the username right? what if i want to validate the username without even clicking the button?
so for example: i have 2 inputs, username and password. in this case i type “rocky” into the username box, and when i click my mouse on the password box (onchange), the website will alert whether the username is available or not. How can i achieve that?
Thank you.

Regards
Rocky

Please remove– $available = true;

and

use the following lines

$result = mysql_query($query);
$rows= mysql_num_rows($result);

then fixed your problem (Error).

Thanks & Regards.

Sava Intl.

So in the above code,
REMOVE – $available = true;

and

ADD the following two lines

$result = mysql_query($query);
$available = mysql_num_rows($result);

That should solve your problem.

Thanks
Sachin Khosla

it doesn’t work. I post my code here.

<?php
$DatabaseHost = "localhost";
$DatabaseUser = "xxx";
$DatabasePassword = "xxx";
$Database = "db";

mysql_connect($DatabaseHost, $DatabaseUser, $DatabasePassword) or die(mysql_error());
mysql_select_db(db);

$user = strip_tags(trim($_REQUEST['username']));

if(strlen($user) -1,
‘result’ => ‘Invalid username, please try again.’
));
die;
}

$query = “SELECT * FROM userdata WHERE Nickname = ‘$user’ “;

$available = true;

if($available)
{
echo json_encode(array(‘code’ => 1,
‘result’ => “Success,username $user is still available.”
));
die;
}
else
{
echo json_encode(array(‘code’ => 0,
‘result’ => “Sorry but username $user is already taken.”
));
die;
}
die;
?>

When i validate the nickname ever come Success,username xxx is still available

Whats wrong?
Pls help me.

Thanks
Janine

The mysql query depends upon your database structure, i.e. table name and the column names. The simplest query would be,

$query = “Select * from USER_TABLE where username = ‘$user’”;

You need to replace
USER_TABLE with the table name in whcih users are stored.
username with the actual column name you have in your table.

If you still have any problem, feel free to ask.

Thanks
Sachin Khosla

I have a problem with this script. I will check the usernames from a mysql database.

How can i read the usernames from the database?

Pls help me

Janine