return value from getJSON function

Question

I have a function with jquery getJSON and i need the return the result value back (to Use it somewhere else)

Here is the code:

function getval(){
jQuery.getJSON('http://data.mtgox.com/api/1/BTCUSD/ticker?callback=?', function(data) {
    // We can't use .return because return is a JavaScript keyword.
    return data['return'].avg.value;
});
}

$(function () {
    $(document).ready(function() {
    alert (getval());
    });

});

This is doesn't work :(

i know i can call external function from inside the getJSON function with the value like:

    jQuery.getJSON('http://data.mtgox.com/api/1/BTCUSD/ticker?callback=?', function(data) {
        // We can't use return because return is a JavaScript keyword.
       mysecondfunction(data['return'].avg.value);
    });
function mysecondfunction(value){
//use the value
}

But i have to call the json function from another function because json return a dynamic value and i need to use it.

I hope it clear...

Thank you very much!!

ajax is asynchronous, anything that needs the result needs to be ran in a callback that gets called after the ajax call completes. — user400654
– user400654, Commented Apr 15, 2013 at 17:18
possible duplicate of How to return a value from a function that calls $.getJSON? — user229044
– user229044 ♦, Commented Apr 15, 2013 at 17:21
possible duplicate of How to return the response from an AJAX call? — Bergi
– Bergi, Commented Apr 30, 2014 at 21:22

Eran Levi · Accepted Answer · 2013-04-16 08:38:34Z

9

Here is the final solution:

function getval( callback ){
    jQuery.getJSON('http://data.mtgox.com/api/1/BTCUSD/ticker', function(data) {
        // We can't use .return because return is a JavaScript keyword.
        callback(data['return'].avg.value);
    });
}

$(function () {
        $(document).ready(function() {
        getval( function ( value ) { 
            alert( 'Do something with ' + value + ' here!' );
        } );
    });

});

Thanks everyone for your help!!

answered Apr 16, 2013 at 8:38

Eran Levi

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Comments

Kevin Boucher · Accepted Answer · 2013-04-15 18:17:23Z

3

You might try using a callback function:

function getval( callback ){
    jQuery.getJSON('http://data.mtgox.com/api/1/BTCUSD/ticker?callback=?', function(data) {
        // We can't use .return because return is a JavaScript keyword.
        callback(data['return'].avg.value);
    });
}

$(function () {
        $(document).ready(function() {
        getval( function ( value ) { alert( 'Do something with ' + value + ' here!' ) } );
    });

});

edited Apr 15, 2013 at 18:17

answered Apr 15, 2013 at 17:17

Kevin Boucher

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4 Comments

Eran Levi Over a year ago

Hi! just a small question.. when i use: "alert (getval( function ( value ) { alert( 'Do something here!' ) } ));" i get "undefined" can you see why?? Thank you very much!

John S Over a year ago

@EranLevi - I think Kevin didn't mean for that first "alert" to be there. The line should be more like this: getVal(function(value) { alert('value is: ' + value); }.

Eran Levi Over a year ago

Thank you! you probably meant to write: getVal(function(value) { alert('value is: ' + value); }); but the page is blank now :( can you see anything else?? Thanks man!

Eran Levi Over a year ago

@KevinBoucher - Hello and thank you!! i still can't see what is wrong in this simple code: link Thank you!

John S · Accepted Answer · 2013-04-15 17:15:24Z

1

Ajax calls are asynchronous, so you can't have the getVal() function return something. Whatever you need to do with the result, you have to do it in inside the callback function.

function getval() {
    jQuery.getJSON('http://data.mtgox.com/api/1/BTCUSD/ticker?callback=?', function(data) {
        // You have to use "data" here
        alert(data['return'].avg.value);
    });
}

$(function () {
    $(document).ready(function() {
        getval();
    });
});

answered Apr 15, 2013 at 17:15

John S

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4 Comments

Eran Levi Over a year ago

You right, but what if i need to refresh (get) the json call again and again? (json array values changing dynamically!) thank you!

user400654 Over a year ago

@EranLevi Then send the request again.

Eran Levi Over a year ago

first.. Thanks! now, i can't send the request beause im calling from function No.1 to Function No.2, now here i cant call Function No.1 again (because it will call Func.No.2... :)... ) i really need to use the return value somehow in function No.2..

John S Over a year ago

@EranLevi - You will have to move some of the code from the the one function to the other so it is inside the ajax callback function. I'd have to see the functions to comment further.

Kathir · Accepted Answer · 2014-04-10 12:07:06Z

1

Hi getJSON asynchronous call so its return undefind

so you need to fire ajax call pass with this args asnyc:false

Ex:

    function getCountrycodeJson(obj) {
     var code="";
        $.ajax({
         async: false,
          dataType : 'json',
         url: "url",
         type : 'GET',
         success: function(data) {
         for(var i in data){ 
//here do your logic and assign value for code varable   

          }
           }
      }});

        return code;
    }

this is working for me.....

edited Apr 10, 2014 at 12:07

answered Apr 10, 2014 at 12:01

Kathir

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