Javascript variable gives undefined error?

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I have following code:

JavaScript

var employee = '';
       function FillEmployee()
       {
           $.ajax({
               type: "POST",
               url: 'http://localhost:1732/GridViewExample/ClientServerInterface.aspx/GetEmployee',
               data: "{}",
               contentType: "application/json; charset=utf-8",
               dataType: "json",
               async: false,
               success: function (id)
               {

                     employee = id.d.employee;
                    PopulateValueDropDown(employee);
               }
           });
       }

in above code variable employee gives undefined error but i have already globaaly defined.

Posted 6-Dec-12 5:37am

Sandeep Shekhawat

Updated 6-Dec-12 5:54am

AmitGajjar

v2

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Comments

BC @ CV 6-Dec-12 11:45am

Is it possible that id.d.employee is giving you the undefined error? Are you sure you're getting back the data you expect from the ajax call?

Sandeep Shekhawat 6-Dec-12 12:01pm

Yes, It is returning data. I have debug code and checked it.

BC @ CV 6-Dec-12 12:30pm

So if you are returning data from the ajax call what does parameter id contain?

ZurdoDev 6-Dec-12 12:22pm

If employee is undefined then id.d.employee must be undefined.

Adam R Harris 6-Dec-12 13:21pm

Using a javascript debugger (IE Developer Tools), set a break point and add a watch for id.d.employee and employee then see what the value coming back from the ajax call is and if its getting set properly. All the other comments are correct, looking at the code the only way employee would be undefined is because id.d.employee is undefined.

You are initializing the employee variable to '' so the only possible way it could be undefined is if id.d.employee is undefined or if it is getting set to undefined somewhere else in your code. After you set your break point double check the property name as i'm fairly certain its just a typo that is causing your problem.

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ss29101976 · Answer 1 · 2012-12-06T09:22:00

Solution 1

change variable name

var employee = '';

to some other name

var emp = '';

and use that variable in the code.

Posted 6-Dec-12 9:22am

ss29101976