CODEWITHSUNDEEP
javascript
A. Z.
A. Z.

Reputation: 736

Run one function after another: callback issue

I'm trying to understand why function lazy_2 doesn't run this way:

function lazy_1 (callback) {
    alert('lazy_1');
}

function lazy_2 () {
    alert('lazy_2');   
}

lazy_1(lazy_2);

? (Via Execute jquery function after another function completes.)

jQuery method works well:

function lazy_1 (callback) {
    alert('lazy_1');
}

function lazy_2 () {
    alert('lazy_2');   
}

$.when( lazy_1() ).done(function() {
    lazy_2();
});

http://jsfiddle.net/5LL69/

Upvotes: 0

Views: 1120

Answers (2)

jfriend00
jfriend00

Reputation: 707466

Because lazy_1() doesn't call it's callback - in fact nobody does. It needs to look like this for the callback to get called:

function lazy_1 (callback) {
    alert('lazy_1');
    callback();
}

function lazy_2 () {
    alert('lazy_2');   
}

lazy_1(lazy_2);

Your second code block above is equivalent to:

lazy1();
lazy2();

because you're just asking jQuery $.when() to run one function and then another with no promises involved (all synchronous code).

Upvotes: 4

Paul Draper
Paul Draper

Reputation: 83273

You have to actually call callback.

function lazy_1 (callback) {
    alert('lazy_1');
    callback();
}

function lazy_2 () {
    alert('lazy_2');   
}

lazy_1(lazy_2);

Upvotes: 2

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