为什么我不能访问JS对象属性？

Question

I have the following function:

function getAggregateData(){
        var sums = new Object();


    $.getJSON("example.json.php", function(data) {
           //for each month
           c = 0;
            $.each(data, function(key, val, index) {
                  //for each store
                $.each(val, function(key2, val2, index2) { 
                      if(c == 0){
                         sums[key2] = val2; 
                      }
                      else{
                         sums[key2] += val2; 
                      }

                });
                c++
            });

    })
    return sums;
}

Which I then call as such:

var totals = getAggregateData();

But when I console log I am totally stumped:

console.log(totals)

reveals an object like this:

    store1  500
    store2  900
    store3  750
    and so on and so forth...

but when I do console.log(totals['store1') I get undefinded.

I have also tried console.log(totals.store1)

and console.log(totals[0].store1)

I am having some type of issue of scope, or I am not creating the object I think I am.

Answer 1

It looks like the function would be returning an empty object since it's not waiting for the AJAX call to finish.

If you tried doing console.log(totals.store1) on the last line inside your $.getJSON callback you'll probably get a result.

You'll need to put any code that requires data from "example.json.php" inside a callback that only gets run after the AJAX call has returned.

Eg

function getAggregateData(){
    var sums = new Object();

    $.getJSON("example.json.php", function(data) {
       //for each month
       c = 0;
        $.each(data, function(key, val, index) {
              //for each store
            $.each(val, function(key2, val2, index2) { 
                  if(c == 0){
                     sums[key2] = val2; 
                  }
                  else{
                     sums[key2] += val2; 
                  }

            });
            c++
        });

        processAggregateData(sums);

    })
}

function processAggregateData(totals) {
    console.log(totals.store1);
}

getAggregateData();

Answer 2

given:

{
    "1": {
        "store1": 2450,
        "store2": 1060,
        "store3": 310
    },
    "2": {
        "store1": 2460,
        "store2": 1760,
        "store3": 810
    }
};

This should work if you intend to add the results for each store.

/**
* This functions need to be called when we have the data
*/
function processSums(obj){
    console.log(obj);
}

function getAggregateData(){

    var sums = {};

    $.getJSON("example.json.php", function(data) {
            $.each(data, function() {
                $.each(this, function(key, val, index){                
                    sums[key] = sums[key] || 0;
                    sums[key] += val;               
                });
            });
            // 4910
            processSums(sums);
    });

    return sums;
}

getAggregateData();