Run something only once, not using .one ()

Question

I've code a piece of code that goes like this:

$('img').load(function(){
    $(this).doStuff(); // here could be one, two, or more images
    doSomethingElse(); // I want this part to run only ONCE
});
function doSomethingElse(){
    // code here unfortunately runs one, two, or more times (depending on how many imgs there are)
}

I'm not sure how to apply jQuery .one() in this situation. I've tried searching online for a way to run something only once, but all I ever seem to find are recommendations to try using jQuery .one().

There must be some way to run doSomethingElse only once, even if multiple images are loaded. I suppose I could use a setTimeout and wait for all the images to have loaded, but I would be guessing how long this takes so that solution seems a bit weak.

You can look into [Self Invoking Function](http://stackoverflow.com/questions/18203897/a-self-invoking-anonymous-function-expression). Basic syntax: `(function(){})()`. These function will execute only once when file is loaded. — Rajesh, Oct 31 '15 at 17:56
try this `$('img').load(function(){ $(this).doStuff(); // here could be one, two, or more images doSomethingElse(); // I want this part to run only ONCE return; // make it run only once });` — Ari, Oct 31 '15 at 18:09

score 1 · Accepted Answer · answered Oct 31 '15 at 17:10

You could accomplish this using a simple boolean flag:

var done = false;

$('img').load(function(){
    $(this).doStuff(); // here could be one, two, or more images

    if (!done)
    {
        done = true;
        doSomethingElse(); // I want this part to run only ONCE
    }
});

function doSomethingElse(){
    // code here unfortunately runs one, two, or more times (depending on how many imgs there are)
}

This way, you set your done variable the first time the load function runs, and each time it runs thereafter, it skips the call to doSomethingElse().

Run something only once, not using .one ()

1 Answers1