Comments on: Json, Jquery, $.ajax(), asp.net and c# to get ajaxed data table rows passing multiple parameters

By: cris_gomez

cris_gomez — Mon, 24 May 2010 08:55:50 +0000

The code below is not advisable since this is prone
to sql injection.
string strFind = String.Format(“FirstName like ‘{0}%’ AND LastName like ‘{1}%’ “, firstName.Trim(), lastName.Trim());

By: salman

salman — Fri, 30 Apr 2010 10:26:33 +0000

Very Helpful article…

By: nicholas

nicholas — Fri, 28 Aug 2009 09:31:28 +0000

Hi Naveed，
‘var myObject = eval(’(’ + msg.d + ‘)’);’ also worked for me. But finally I decided to use parse2.js instead of eval()..
If you never use this json converter file, maybe you could try to use it.

Thanks for your excellent article.

By: Prawin

Prawin — Thu, 27 Aug 2009 07:36:58 +0000

Atlast it worked for me.

In the http://www.javascriptkata.com/2007/05/08/how-to-use-json/#comment-710
eval(”(” + eval(theServerSideJsonTextResponse); + “)”) did your job..but it didn’t worked for me..

I have used the following pice of code which worked exactly.
var myObject = eval(‘(‘ + msg.d + ‘)’);

Anywayss thanks for the article.

By: Prawin

Prawin — Wed, 26 Aug 2009 13:47:56 +0000

Even Iam getting the same error. Iam using Firefox. Please help me out.
Prawin

By: Naveed

Naveed — Tue, 25 Aug 2009 15:40:47 +0000

You can try

var myObject = eval(’(’ + myJSONtext.d + ‘)’);

By: nicholas

nicholas — Mon, 24 Aug 2009 20:29:02 +0000

Hi,
I tried to use var myObject = eval(‘(‘ + myJSONtext + ‘)’); for json convert to javascript, but it not work for me. can you help me ?
the error message: ‘]’ expect

Thanks

By: Jakub Konecki

Jakub Konecki — Mon, 24 Aug 2009 15:07:27 +0000

You should read about SQL Injection before you release the code from this article.