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

Question

This question already has answers here:

How do I return the response from an asynchronous call? (42 answers)

Closed 9 years ago.

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');

Answer 1

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);

Answer 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

Answer 3

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');

Answer 4

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
});