Making sure a javascript function is not called, if called already

Question

I am using google maps for a project. I need to be able to hide and display multiple maps. I cannot use a basic toggleDiv type function alone. This is because Google maps will ignore the intended size of the div when the div is set to display: none from CSS. (for whatever reason it is fine with being toggled by javascript.) I could use GSize(width, height) but it cannot handle percentages. Since I need the map to be 100%, 100%, this is not an option. I figured out a way around it which is to call the second map's function using onClick, rather then loading all the functions using body onLoad. But, then the zoom of the map is not saved and the map is just reloaded.

So, I need to check if a function has been called, and if so, do not recall it. I cannot figure out how to do this. Any help is appreciated.

Thank you.

For a prettier solution - see teresko's answer. Tracking variables are kept INTERNAL to the function and don't clutter the outer scope. — John Strickler
– John Strickler, Commented Jul 6, 2011 at 20:33

katspaugh · Accepted Answer · 2011-07-06 20:54:55Z

7

Another "lazy" variant:

var doOnce = function () {
    // do something

    doOnce = function () {}
}

As for the original problem, if the function you want to run once is the click event handler, you'd better remove the click event listener.

edited Jul 6, 2011 at 20:54

answered Jul 6, 2011 at 20:32

katspaugh

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Comments

brenjt · Accepted Answer · 2011-07-06 20:14:41Z

2

Define a variable as such. var beenFired = false;

in the function do this:

function myFunction()
{
    if(!beenFired)
    {
        //TODO FUNCTION
        beenFired = true;
    } 
}

That will check to see if hasn't been fired yet and if not fire it. You could even do it around the function call itself.

if(!beenFired)
    myFunction()

function myFunction()
{
    //TODO FUNCTION
    beenFired = true;
}

answered Jul 6, 2011 at 20:14

brenjt

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4 Comments

Mitchell Ingram Over a year ago

I literally just figured it out. I tried to answer my own question but it wouldn't let me. Thanks!

jfriend00 Over a year ago

I believe you're allowed to answer your own question, but only after some time period has passed. You can always edit your original question to include additional discoveries.

brenjt Over a year ago

Yes you able to so, just as @jfriend00 said you'll have to wait a certain amount of time.

Mitchell Ingram Over a year ago

Yeah, I was just going to edit it but saw so many people answered it didn't matter. Sometimes you just need to talk it out before you get that thud! moment. Appreciate all the responses.

tereško · Accepted Answer · 2011-07-06 20:21:54Z

1

var func = (function(){
    var first = true;
    return function( a, b, c){
        if ( !first ){
            return;
        }
        first = false;

        console.log( arguments );
    }
})();

func( 1, 2, 3); // in console : [ 1, 2, 3]
func( 1, 1, 1); // nothing happens;

answered Jul 6, 2011 at 20:21

tereško

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1 Comment

katspaugh Over a year ago

It's a waste of computation (the condition is evaluated on every call), see my answer below.

Sergey Metlov · Accepted Answer · 2011-07-06 20:58:13Z

1

If I understand:

var isCalled = false;

function f() {
  if (!isCalled) {
    isCalled = true;
    // ...
  }
}

edited Jul 6, 2011 at 20:58

answered Jul 6, 2011 at 20:16

Sergey Metlov

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5 Comments

Sergey Metlov Over a year ago

Oh, first I thought, that it is necessary to keep only 1 working function at single moment. Thanks, I've fixed.

katspaugh Over a year ago

At single moment? You mean during an Ajax request?

Sergey Metlov Over a year ago

I mean to prevent second calling of the function while the first one is running. It is often situation for some event handlers. But I've read question carefully and fixed my answer.

katspaugh Over a year ago

Just making sure you understand that there can't be simultaneous calls in synchronous environment (which JavaScript is).

Sergey Metlov Over a year ago

Yea, I understand. JS is a sync. But it could be an ajax call in the func body, as you told in 3rd comment.

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