Callback issue with jquery/js/ajax

Question

UPDATE3 and FINAL: SOLVED thanks to Evan and meder!

UPDATE2: I should clarify, I need function updateFilters (a,b) to be called, not created. The function exists elsewhere. Sorry for confusion.

The code below does not work as expected - udpateFilters(a,b) is being called before for loop ends. Changing async to false solves the problem, but this does not seem right. Moreover updateFilters() is called regardless of ajax success.

What did I do wrong?

save (updateFilters(a,b));

function save (cb) {
    $.ajax ({
         url:'save.php',
         data:somedata,
         async:true,
         success: function (response) {
             for (var i in response) {}

             if (typeof cb!='undefined') cb;
    });
}

function updateFilters (a,b) {
//do some stuff here
}

ANSWER: So as meder suggested I change the parameter. This is 2/3 of the answer!

save (
    (function (a,b) {
        return function () {
            return updateFilters(a,b);
        }
    })(a,b)
);

Now change the following:

if (typeof cb!='undefined') cb (); //brackets are important!

And good to go!

meder omuraliev · Accepted Answer · 2010-10-10 22:40:47Z

4

You are passing the invocation of a function, not a function that should be called at a later point. If you want the latter, do this:

save (
   (function(a,b){
      return function() {
        return updateFilters(a,b)
      }
   })(a,b)
);

answered Oct 10, 2010 at 22:40

meder omuraliev

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6 Comments

selytch Over a year ago

Thank you! I'm trying to understand the sequence of calls, which function is called when. Can you help?

meder omuraliev Over a year ago

Put alerts inside of each function with unique values and when you see those alerts you'll know which functions are called.

selytch Over a year ago

that's my issue. alert from the second return is never called.

selytch Over a year ago

i mean putting alert after return function (){alert ('r1');... never appears

meder omuraliev Over a year ago

You are now referencing code that no one is able to see. Please post the modified alert code you have.

|

Evan Trimboli · Accepted Answer · 2010-10-11 00:16:06Z

1

Regarding your update, you never invoke the callback. It's the same as doing

x; // does nothing

You want to call the callback:

cb(); // invoke the function

answered Oct 11, 2010 at 0:16

Evan Trimboli

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1 Comment

selytch Over a year ago

Thanks! I did not realize importance of parentheses!

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