Failed to change a global variable in JQuery ajax function

Question

 $( document ).ready(function() {
        function doAjax( time_from, time_to ){
            var dataRsp;
            $.ajax({
              url: "/query/"+time_from+"/"+time_to,
              type: "GET",
              dataType: "json",
              success: function(data){ dataRsp = data; },
            });
            alert(JSON.stringify(dataRsp));
        };
      doAjax(0,0);
  }

The above is my code snippet, I need to store the ajax response datain a global variable dataRsp, but I failed to do this.I am very confuse with the variable scope in JS and jquery.Thank you very much.

You didn't fail to change it, it's just asynchronous ! — adeneo
– adeneo, Commented Mar 28, 2014 at 6:15
possible duplicate of How to return the response from an AJAX call? — adeneo
– adeneo, Commented Mar 28, 2014 at 6:16

iJade · Accepted Answer · 2014-03-28 06:32:52Z

3

Put your alert inside the success callback

    $( document ).ready(function() {
        function doAjax( time_from, time_to ){
            var dataRsp;
            $.ajax({
              async: false,
              url: "/query/"+time_from+"/"+time_to,
              type: "GET",
              dataType: "json",
              success: function(data){ 
                  dataRsp = data; 
                  return(JSON.stringify(dataRsp)); 
              }
            });

        };
      var x =doAjax(0,0);
      alert(x);
  }

Or Another option would be to add async: false parameter. And also the , after success is not required.

edited Mar 28, 2014 at 6:32

answered Mar 28, 2014 at 6:16

iJade

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1 Comment

stackpop Over a year ago

How to change it if I want the function doAjax return dataRsp, and I need to use the data outside the function.for example, I write x = doAjax(0,0), and then alert x.

yarmyarch · Accepted Answer · 2014-03-28 06:25:08Z

0

// GLOBAL
var dataRsp;
$( document ).ready(function() {
    function doAjax( time_from, time_to ){
        // var dataRsp;
        $.ajax({
          url: "/query/"+time_from+"/"+time_to,
          type: "GET",
          dataType: "json",
          success: function(data){ 
              dataRsp = data; 
              // YOU GET IT HERE
              alert(JSON.stringify(dataRsp)); 
          },
          // Add this if you want to get it just after the "ajax" but not in the callback of "success"
          async: false
        });
        // ALWAYS NULL if async==true as it's not returned yet.
        // alert(JSON.stringify(dataRsp)); 
    };
  doAjax(0,0);

}

answered Mar 28, 2014 at 6:25

yarmyarch

1

1 Comment

Nevin Over a year ago

Please do provide more details about your answer.

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