CODEWITHSUNDEEP
javascriptjquery
jamone
jamone

Reputation: 17421

Why is this javascript function running without being called?

$(document).ready(SetupButtonClicks());

function SetupButtonClicks() {
    $('#btnJavaPHP').click(DoPHPStuff());
}

function DoPHPStuff() {
    //stuff
}

I have this code in my javascript file, when I debug it I see it call SetupButtonClicks() like it should but after that is done it calls DoPHPStuff(). DoPHPStuff() should only be called when btnJavaPHP is clicked. What am I doing wrong?

Upvotes: 5

Views: 5230

Answers (4)

Pointy
Pointy

Reputation: 413709

Change your SetupButtonClicks function:

$('#btnJavaPHP').click(DoHPStuff);

The way you've got it coded, you're telling Javascript to call the function, not to use it as the "click" handler. The parentheses are an operator that causes a function to be called.

Upvotes: 17

Jeremy
Jeremy

Reputation: 792

With the exception of a function declaration, a pair of parentheses following a function's identifier causes the function to execute. Examples:

// function declaration; function not executed
function SetupButtonClicks() {

}

// function executed
SetupButtonClicks();

// function not executed
SetupButtonClicks;

Upvotes: 2

gblazex
gblazex

Reputation: 50109

function DoPHPStuff() {
    //stuff
}

function SetupButtonClicks() {
    $('#btnJavaPHP').click(DoPHPStuff);
}

$(document).ready(SetupButtonClicks);

Upvotes: 3

SLaks
SLaks

Reputation: 887413

Remove the ().

By writing $(document).ready(SetupButtonClicks()), you are calling SetupButtonClicks and passing its return value to ready.
Similarly, by writing $('#btnJavaPHP').click(DoPHPStuff()), you are calling DoPHPStuff (immediately) and passing whatever it returns to click().

You need to pass the functions themselves by writing $(document).ready(SetupButtonClicks) and $('#btnJavaPHP').click(DoPHPStuff).

Upvotes: 3

Related Questions