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Home/ Questions/Q 8365269
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T12:37:11+00:00 2026-06-09T12:37:11+00:00

I have a circumstance where I need to make several getJSON calls, and once

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I have a circumstance where I need to make several getJSON calls, and once all of the data is returned make a call to another function, like so (code simplified for example):

var data = {};
for (var i in a) {
    $.getJSON(base_url+i, function(d) {
        data[i] = d;
    });
}
do_something(data);

Obviously this doesn’t work as I am making the call to do_something before the getJSON calls have returned any data.

My current approach to get around this is to make the calls synchronously, like so:

var data = {};
$.ajaxSetup({'async':false});
for (var i in funcs) {
    $.getJSON(base_url+i, function(d) {
        data[i] = d;
    });
}
$.ajaxSetup({'async':true});
do_something(data);

My question is, is there a better way of doing this or am I best off making the calls synchronously as above?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T12:37:12+00:00Added an answer on June 9, 2026 at 12:37 pm

    As per link posted by Felix Kling to a similar question, the answer was to use deferred objects.

    However, there is a further complication due to the use of i in the getJSON success function. This would always be the value of the last iteration. See my other question for more details:
    jQuery Deferred – Variable scope in deferred getJSON success functions

    Full solution:

    var data = {};
    var calls = [];
    for (var i in funcs) {
        calls.push(
            $.getJSON(base_url+i,
                (function(i) {
                    return function(d) {
                        data[i] = d;
                    };
                }(i)) 
            )
        );
    }
    $.when.apply($,calls).then(function() {
        do_something(data);
    });
    
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