Comments on: AJAX with jQuery: A Simple Login Example

By: Swetz

Swetz — Wed, 10 Feb 2010 08:55:47 +0000

just for your reference i have this code which is showing error

$(document).ready(function(){
$(‘#error’).hide();
$(‘form[name=loginform]‘).submit(function(){
$(‘#error’).slideUp();
$.post(‘/ajax.php’),{username: $(‘[name=username]‘).val(), password: $(‘[name=password]‘).val()}, function(data) {
if(data.success)
{
location.href = data.redirect;
}
else
{
$(‘#error’).html(data.message).slideDown();
}
}, ‘json’);
return false;
});
});

By: Swetz

Swetz — Wed, 10 Feb 2010 08:47:11 +0000

hey very nice tut.. i wrote the code same like urs.. but im getting this error in firebug

missing ; before statement
}, ‘json’);

m very new to jquery.. so not sure wat its telling me to do. can u plz help?

thanks a lot

By: Shawn McCool

Shawn McCool — Sun, 07 Feb 2010 17:57:18 +0000

@Johnny

I’ll be adding the code to the post shortly.

By: Johnny

Johnny — Sun, 07 Feb 2010 06:03:20 +0000

wish the code was available this is what im looking for.. but stopping and typing out the code is going to be a pain.

i also need to convert this to a mysql check too

thanks for the tut tho

By: indra

indra — Fri, 22 Jan 2010 16:42:55 +0000

veryyyyyyy usefull thank, you so awesome

By: Diraj

Diraj — Wed, 06 Jan 2010 16:41:22 +0000

Great job!….I have gone through some many tutorials before but I dint get anything out of them, but this one helped me to understand step by step..

By: widi dwi

widi dwi — Wed, 30 Dec 2009 12:10:55 +0000

wow …. this tutorial is help me alot ,thx. i`ll wait for another tutorial ..

By: Amy Crompton

Amy Crompton — Sun, 06 Dec 2009 15:35:58 +0000

Great example that degrades very well.

Thanks

By: Mr. Haynes

Mr. Haynes — Thu, 26 Nov 2009 15:47:45 +0000

I have tried several times but I cannot get the error message piece to work, it just does not show up?

Username:
Password:

By: Shawn McCool

Shawn McCool — Sun, 08 Nov 2009 20:34:11 +0000

@Gagan

You need to make sure that your ajax_login.php is only returning a json object, no additional data. One good way to do this would be to use the Firefox extension “Firebug”. You can open the firebug console and enable it for your page. Then, when you try to execute your script you’ll see the specific call it makes to ajax_login.php. You’ll be able to see both what is sent (the request) and the response. The response must be a valid json object. I notice that you’re ‘echo’ing quite a few extra values. I understand that this may have been debug code, but it will break jQuery’s interpretation of the results.