CODEWITHSUNDEEP
jqueryajax
user1883013
user1883013

Reputation: 1

onblur username availability checking

I am doing a username availablity on blur using jquery ajax

respose is Available getting in firebug

My problem is the loading image is keep on rotating.

ReferenceError in firebug console: finishAjax is not defined

setTimeout("finishAjax('Info', '"+escape(response)+"')", 450); Here's it

<script type="text/javascript" src="jquery-1.3.2.js"></script>
<link href="css.css" media="screen" rel="stylesheet" type="text/css" />

<script type="text/javascript">

function check_username() {

$.post("check_username_availablity.php", {
username: $('#name').val(),
}, function(response){
//$('#Info').fadeOut();
//$('#Loading').hide();
setTimeout("finishAjax('Info', '"+escape(response)+"')", 450);
});

function finishAjax(id, response) {

$('#'+id).html(unescape(response));
//$('#'+id).fadeIn(1000);
}

}
</script>

</head>
<?php
include('dbcon.php');?>
<body>

<form action="#" name="customForm" id="customForm" method="post" enctype="multipart/form-data">

<div class="both">
<h4> Try "Zeeshan" , "John" , "Katty", "Jimmy" </h4><br clear="all" /><br clear="all" />
<br clear="all" />
<div>
<label>User Name</label>
<input id="name" name="username" type="text" value="" onblur="check_username()"/>
<div id="Info"></div>
<span id="Loading"><img src="loader.gif" alt="" /></span>
</div>
</div>

</form>

</body>
</html>


<?php
include('dbcon.php');
if($_REQUEST)
{
$username   = $_REQUEST['username'];
$query = "select * from register where username = '".strtolower($username)."'";
$results = mysql_query( $query) or die('ok');

if(mysql_num_rows(@$results) > 0) // not available
{
echo '<div id="Error">Already Taken</div>';
}
else
{
echo '<div id="Success">Available</div>';
}

}?>

Thanks..

Upvotes: 0

Views: 1142

Answers (2)

Adil
Adil

Reputation: 148110

You need to pass function name to setTimeout not call it, or you can use anonymous function to call function it its body.

setTimeout(function(){
     finishAjax('Info', escape(response))
}
, 450);​

Upvotes: 1

user1726343
user1726343

Reputation:

If you're using a setTimeout, your function needs to close over the response variable. Try this:

function(response){
    setTimeout(function(){
        finishAjax('Info', escape(response));
    }, 
    450);
});

Upvotes: 1

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