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Home/ Questions/Q 7534271
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T05:58:07+00:00 2026-05-30T05:58:07+00:00

I want to change the background image of a button using Javascript. Here is

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I want to change the background image of a button using Javascript. Here is what I am currently trying, but it is failing.

HTML code –

      <tr id="Rank1">
              <td><button class="darkSquare"></button></td>
               <td><button class="lightSquare" ></button></td>
                <td><button class="darkSquare" ></button></td>
               <td><button class="lightSquare" ></button></td>
                <td><button id="e1" class="darkSquare" ></button></td>
                <td><button class="lightSquare" ></button></td>
                <td><button class="darkSquare" ></button></td>
                <td><button class="lightSquare"> </button></td>
            </tr>

JavaScript Code

        initializer();

        function initializer()
         {
           var tableRow = document.getElementById("Rank1");

           var buttons = tableRow.getElementsByTagName("button");

           for(b in buttons)
            {
               console.log(b.toString());
               b.style.backgroundImage = "url('darkSquare.jpg')";
            }
         }

In the console, I get an error saying b.style is undefined.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T05:58:08+00:00Added an answer on May 30, 2026 at 5:58 am

    for (... in ...) loops do not work that way in JavaScript it should be:

    for (var b = 0; b < buttons.length; b++) {
        buttons[b].style.backgroundImage = "url('darkSquare.jpg')";
    }
    

    for (... in ...) actually iterates over all the “members” of an object

    eg. using var x = {a: 1, b: 2, c: 3} in for(var m in x) console.log(m) will produce

    > a
    > b
    > c
    

    it kind of works with arrays because it considers the indices members like this:

    var x = [1,2,3];
    for (var m in x) console.log(m);
    > 0
    > 1
    > 2
    

    since it is giving you the indices as if they were members you can’t distinguish. the pitfall is:

    var x = [1,2,3];
    x.stuff = 'boo!';
    for (var m in x) console.log(m);
    > 0
    > 1
    > 2
    > stuff
    

    General Rule of Thumb: only use for (... in ...) when iterating over members in an object, use for (var i = 0; i < array.length; i++) when using arrays

    you can always cheat and use:

    for (var i = 0, item = x[i]; i < x.length; item=x[++i]) {
        // now item is the current item
    }
    
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