javascript - change the style of img inside a div tag - Stack Overflow

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Example:

<div class="test"><img src="test.jpg" style="display:none;" /></div>

How to change that with javascript, I know in css its:

.test img {display:block;}

but in javascript I only know this:

document.getElementById("test").style.display="block";

thats obviously the whole div tag , not the img.

Example:

<div class="test"><img src="test.jpg" style="display:none;" /></div>

How to change that with javascript, I know in css its:

.test img {display:block;}

but in javascript I only know this:

document.getElementById("test").style.display="block";

thats obviously the whole div tag , not the img.

• javascript

Share Improve this question Follow asked Nov 24, 2011 at 14:57 N_F_SN_F_S 10922 gold badges22 silver badges99 bronze badges 1

• Your element has a class, not an ID ... – Alnitak Commented Nov 24, 2011 at 15:00

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 7

If you're using an ID "test", you can do this.

document.getElementById("test")
        .getElementsByTagName("img")[0].style.display="block";

This uses getElementsByTagName which returns a collection of the images found. The [0] grabs the image at index 0 (the first image), and then applies the style.

If you have a class "test", you can do this, but it won't work in IE7 and below:

var imgs = document.querySelectorAll(".test > img");

for( var i = 0; i < imgs.length; i++ ) {
    imgs[i].style.display="block";
}

For the widest browser patibility, you can do this:

function getElementsByClassName( root, clss ) {
    var elems = document.getElementsByTagName('*'),
        result;
    clss = " " + clss + " ";
    for( var i = 0; i < elems.length; i++ ) {
        if( (" " + elems[ i ].className + " ").indexOf( clss ) > -1 ) {
            result.push( elems[i] );
        }
    }
    return result;
}

var tests = getElementsByClassName( document, "test" );

for( var i = 0; i < tests.length; i++ ) {
    var img = tests[i].getElementsByTagName('img')[0];
    if( img ) {
         img.style.display="block";
    }
}

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