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Home/ Questions/Q 1008455
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T08:46:24+00:00 2026-05-16T08:46:24+00:00

function json (url){ $.getJSON(url, function(data) { return data; }) } this function don’t see

  • 0
function json (url){
       $.getJSON(url, function(data) {
           return data;
       })
   }

this function don’t see “data”

function MakeAlert(){
       data = json('action.php');
       window.alert(data.name);
}

this work:

  function json (url){
           $.getJSON(url, function(data) {
               window.alert(data.name);
           })
  }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T08:46:25+00:00Added an answer on May 16, 2026 at 8:46 am

    That’s because the $.getJSON is asynchronous. It sends the request and returns immediately. Once the server responds (which might be a few seconds later) it invokes the success callback and that’s why the data is accessible only inside this callback.

    If you want to block the caller you could send a synchronous request (note that this might freeze the UI while the request is executing which defeats the whole purpose of AJAX):

    function json(url) {
        var result = null;
        $.ajax({
            url: url,
            async: false,
            dataType: 'json',
            success: function(data) {
                result = data;
            }
        });
        return result;
    }
    
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