CODEWITHSUNDEEP
jqueryfadeout
Rishi
Rishi

Reputation: 47

Calling a user defined function as callback parameter in fadeOut()

A very basic question:

$(element).fadeOut("slow",myFunction());

This doesn't work. What is the correct way to do it?

Upvotes: 0

Views: 72

Answers (2)

Girish
Girish

Reputation: 12127

Try this way 

$(element).fadeOut("slow", function(){
   myFunction();
});

Upvotes: 0

Alex
Alex

Reputation: 10226

Try

$(element).fadeOut("slow", myFunction);

Your code, i.e. adding () would actually run the code rather than defining it as callback.

Just some quick console playing:

function test() {}

typeof test // "function"
typeof test() // "undefined"

So the argument needs to be a function and not what a function returns (except that would be function, too).

Upvotes: 3

Related Questions