如何从javascript / jQuery中调用另一个函数的函数获取返回值？

Question

So you can assign a value to var doub as below:

function getDouble(number) {
    var doubl = number + number;
    return doubl;
}

var num = 4;
var doub = getDouble(num);

I want to know if I can get the return type by calling a second function as below. So can the return value of the second method be assigned to doub ?

function getDouble(number) {
    var doubl = number + number;
    return quadruple(doubl);
}

function quadruple(doubl) {
    quad = doubl + doubl
    return quad;
}

var num = 4;
var doub = getDouble(num);

For some reason it doesn't work for me.

This is my code with the problem..I am doing something similar to the above, but not working:

In my doc ready, I have this code..I want the result to be stored in newjsn

    $(document).ready(function() 
    {
     var newjsn= DatabaseCategoriesToSelectList(num, 200, 100);

My DatabaseCategoriesToSelectList function is :

function DatabaseCategoriesToSelectList(jsn, max, min){

    var ddl="<option selected='selected' value='"+min+"'>Select...</option>";

    //load categories from database, and put them into the select drop down.
    $.getJSON(GLOBALURL+'content/plugins/subcategories/getSubcategories.php',             
        function(data, ddl) {

            var datas=  jQuery.grep(data, function(element, index){
                return element.category_id < max && element.category_id>min; // retain appropriate elements
            });
            for (x=0; x<datas.length; x++){
                ddl= ddl+"<option value='"+datas[x].category_id+"'data-subcategory='"+datas[x].category_subselect+"'>"+datas[x].category_safe_name+"</option>";
            }   
            alert (ddl); return runafter(ddl, jsn); 
        });

}

function runafter(ddl, jsn){
    jsn=ddl;
    alert(jsn);

    return 500;

}

For some reason, the alerts work, but I can't get it to return me the string. I even tried to return a random value of 500, and it doesnt..

Answer 1

The problem is that you're never returning out of your DatabaseCategoriesToSelectList function. Your function looks like this:

function DatabaseCategoriesToSelectList(jsn, max, min) {
    ...
    $.getJSON(GLOBALURL+'content/plugins/subcategories/getSubcategories.php',             
        function(data, ddl) {
             ...
             return runafter(ddl, jsn); 
        }
    );
}

The inner return is not returning out of DatabaseCategoriesToSelectList , it is returning out of the inner function. However, there is no way to make this return the values that you want no matter what you do, because $.getJSON is an asynchronous call. You have to put something like

function(data, dll) {
    ...
    var newjsn = runafter(ddl, jsn);
    // put the rest of your stuff that uses newjsn here
}

in your code, and then put whatever came after that also into that function also.

Answer 2

Fiddle ( output is : 16 )

What you are doing is just fine as it should

 function getDouble (number){
    var doubl = number+number;
    return quadruple(doubl);
    }

function quadruple(doubl){
var quad=doubl+doubl
return quad;
}

var num = 4;
var doub = getDouble(num);

Answer 3

You aren't going to have access to the return from a getJson handler. The framework calls that handler. You are giving it a function, and .getJson calls that function if/when the response is received. If you need access to the return of runAfter, you need to do whatever with it you plan to.

Instead of:

$.getJSON(GLOBALURL+'content/plugins/subcategories/getSubcategories.php',
  function(data, ddl) {

          ...
           return runafter(ddl, jsn); 
  });
}

You need to decide what you want to do with that value, for example I stuff it into the value of some hidden field maybe:

$.getJSON(GLOBALURL+'content/plugins/subcategories/getSubcategories.php',
  function(data, ddl) {

          ...
          $('#someField').val( runafter(ddl, jsn) ); 
  });
}