JavaScript - Define function without calling it?

Question

I feel so so stupid for forgetting this, but I've been out of practice for a minute, and I'm drawing a blank.

Why is slideDown being called onload rather than when the click is handled?

function buttonClicked(buttonNumber) {
    $contentBox.slideDown("slow");
};

$button1.click = buttonClicked(1);

Because you are calling buttonClicked. The () after a function reference always calls the function. Example: function foo() { alert(42); }; foo();. Here foo is called because I put () after the variable name. Btw, if $button1 is a jQuery object, then you have to pass a function reference to the .click method, not assign a value to it. See learn.jquery.com/events. — Felix Kling
– Felix Kling, Commented Jun 12, 2013 at 16:28

ayyp · Accepted Answer · 2013-06-12 16:29:10Z

9

You would want to structure it as

$button1.click(function() {
    buttonClicked(1);
});

This will make it fire when $button1 is clicked.

answered Jun 12, 2013 at 16:29

ayyp

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Comments

Paul Oliver · Accepted Answer · 2013-06-12 16:30:23Z

0

Try this:

$button1.click(function() {
    buttonClicked(1);
});

See documentation at api.jquery.com

answered Jun 12, 2013 at 16:30

Paul Oliver

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