$.json call jQuery doesn't return a value into a variable

Question

i've got a problem with a $.json jQuery call function, this is my code :

// function to get currency live rates from yahoo api keys

function getRate(from, to) {        
  $.getJSON('http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22'+from+to+'%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback=', function(data) {

    var rate  = data.query.results.rate.Rate; 

    return rate; 

  });

}// end of function

var receive = getRate('USD','EUR');

@mplungjan If only `getJSON` had an A in it. Seems all too many people use it without even knowing it is AJAX. — James Montagne, Mar 29 '14 at 19:38

miushock · Answer 1 · 2014-03-30T01:11:42.910

2

$.getJSON is a asynchronous function, which means the callback will be executed once the data is ready, trying to read the data right after asynchronous call will usually get you an empty result.

What you really need here is to pass the variable you want to populate along with the callback function

edited Mar 30 '14 at 01:11

answered Mar 29 '14 at 19:34

miushock

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Can you give me an example please, sorry i am new to programming – Vasileios Tsakalis Mar 30 '14 at 03:52

score 0 · Answer 2 · answered Mar 29 '14 at 19:38

The A in Ajax stands for Asynchronous and you will get your result in the callback not as a return value

function getRate(from, to) {        
    $.getJSON('http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22'+from+to+'%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback=',
      function(data) {
        var rate  = data.query.results.rate.Rate; 
        // here you process
    });
}// end of function

getRate('USD','EUR');

score 0 · Answer 3 · answered Mar 29 '14 at 20:07

An alternative approach would be to simply return the jqXHR object from the function and then work with it after making the call to the function:

function getRate(from, to) {
    return $.getJSON('http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22'+from+to+'%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback=');
}

getRate('USD','EUR').done(function(data) {
    var receive = data.query.results.rate.Rate; 
    // Do your work here
});

Thank you Tim, the code above is what i need and works like a charm! :) — Vasileios Tsakalis, Mar 30 '14 at 03:32

score 0 · Accepted Answer · answered Mar 30 '14 at 13:06

Finally i found the solution of return a value into a variable with $.ajax call function, check the code above, works perfect for me, btw thank you all guys :)

var rate = [ ]; // Step 1 , we create a new array, usefull because we will store here the return value

// Step 2 , we create our function below

function getRate(from,to) {

//Step 3, the ajax call inside a variable

var ajaxCall = $.ajax({

async:false,
dataType:"json",
type:'GET',
url:'http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.xchange%20where%20pair%20in%20(%22'+from+to+'%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys&callback=',
success: function(data) {
result = data.query.results.rate.Rate;
rate.push(result); //we push the result into the array

} });// ajax call end

if(rate.indexOf()<=0) {

ajaxCall.abort();
return calc;

}

}//function finish here

// and then voilà :)

var returnValue = getRate('USD','EUR');

// now do whatever you want

// for example :

alert(returnValue);

a warning, inside the statement its not return calc; but return rate; — Vasileios Tsakalis, Mar 30 '14 at 13:13

$.json call jQuery doesn't return a value into a variable

4 Answers4