javascript variable is not showing correct value [duplicate]

Question

i have this ajax call function.

function saveData(ip)
{

    $JQ.ajax({

     type: "POST",

     url: "all_actions.php",

     data: 
     {
         page_url:document.URL,
         ip_address:ip 
     },

     success: function(responce)
     {

         if(responce)
         {
             var newtoken;
             newtoken=responce;
             return newtoken;
         }
     }
    });
}

Then i have another function

 function getToken()
{
    var ip=myip
    var mytoken;
    mytoken=saveData(ip);
    alert(mytoken);

}

My token giving undefined in alert.Although if i alert newtoken variable in savedata response it gives correct value in alert box.why if i return that avlue it does not assigned to mytoken.

is it something time delay issue.??

Need your help...

AJAX = Asynchronous Javascript And XML. There are many questions with the same problem here on SO :) — Andreas
– Andreas, Commented Jan 23, 2013 at 11:25

ATOzTOA · Accepted Answer · 2013-01-23 11:46:07Z

1

You cannot return from an asynchronous call.

You have to consume the return data inside the success function. Whatever you are going to do with token, write that code inside the success handler.

success: function(responce)
 {

     if(responce)
     {
         var newtoken;
         newtoken=responce;

         // Global variable
         sourceid = newtoken;

         return newtoken; // This won't work
     }
 }

Also

 function getToken()
{
    var ip=myip
    var mytoken;
    mytoken=saveData(ip); // This won't return any data
    alert(mytoken); // This won't give you anything useful

}

edited Jan 23, 2013 at 11:46

answered Jan 23, 2013 at 11:27

ATOzTOA

36.1k23 gold badges101 silver badges119 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Rana.Asif Over a year ago

I have to assigned that value below the page to another javascript variable.thats why i want to return this..like this sourceid = getToken();

ATOzTOA Over a year ago

Then you will have to assign sourceid = newtoken; inside the success handler. Note that sourceid should be global.

Rana.Asif · Accepted Answer · 2013-01-23 11:48:09Z

1

Hi friends This is solution that for i was looking.

    function saveData(ip)
{

    return $JQ.ajax({

     type: "POST",

     url: "all_actions.php",

     data: 
     {
         page_url:document.URL,
         ip_address:ip 
     },
     async: false,

    }).responseText;
}
function getToken()
{
    var ip=myip
    var mytoken;
    mytoken=saveData(ip);
    return mytoken;
}

answered Jan 23, 2013 at 11:48

Rana.Asif

4853 silver badges10 bronze badges

Comments

BenM · Accepted Answer · 2013-01-23 11:26:01Z

0

The first 'A' in AJAX is 'Asynchronous'. Your alert is running before the AJAX request has had a chance to complete. You need to handle whatever you wish to do with the response inside of the success: function() or .done() functions of jQuery:

success: function(responce)
{
    if(responce)
    {
        var newtoken = responce;

        // Work here with newtoken...
    }
 }

answered Jan 23, 2013 at 11:26

BenM

53.3k26 gold badges116 silver badges172 bronze badges

Collectives™ on Stack Overflow

javascript variable is not showing correct value [duplicate]

3 Answers 3

2 Comments

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Linked

Related