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Home/ Questions/Q 7021937
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T23:30:54+00:00 2026-05-27T23:30:54+00:00

I want to pass id1, id2 and id3 through the function below. This works

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I want to pass id1, id2 and id3 through the function below. This works just fine:

function doSomething(id1,id2,id3) {
    $(id1).fadeIn('slow',.25);
    $(id1).fadeIn('slow',.25);
    $(id1).fadeIn('slow',.25);
};

But this does not work:

function doSomething(id1,id2,id3) {
    setTimeout( " $(id1).fadeIn('slow',.25) ", 300);
    setTimeout( " $(id1).fadeIn('slow',.25) ", 300);
    setTimeout( " $(id1).fadeIn('slow',.25) ", 300);
};

How do I get the second one to work? My thought is I need some punctuation around the id’s. Or perhaps I can set a variable for the function within the setTimeout brackets. Any ideas?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T23:30:55+00:00Added an answer on May 27, 2026 at 11:30 pm

    It’s bad practice to get setTimeout() to evaluate a string. Instead, use an anonymous function:

    function doSomething(id1, id2, id3) {
        setTimeout(function() {
            $(id1).fadeIn('slow', 0.25);
            $(id2).fadeIn('slow', 0.25);
            $(id1).fadeIn('slow', 0.25);
        }, 300);
    };
    

    Note that I’ve put all your fadeIn()s into one setTimeout(); it does the same thing as all your timeouts will trigger at the same time (300ms). If you IDs are strings, you could do this too:

    function doSomething(id1, id2, id3) {
        setTimeout(function() {
            $(id1 + ', ' + id2 + ', ' + id3).fadeIn('slow', 0.25);
        }, 300);
    };
    

    Although it’s a little messy, but $.add() might do the trick.

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