Question

Is it possible to do an if statement in javascript/jquery to determine:

If the value of a certain element’s css attribute is equal to a given value?

such as:

if( $('.box').css(background-color = blue) ){
    // do this...
}

Answer 1

The css JQuery method, when given only one parameter, will return the value for the given paramenter. You can try this:

var color = $('.box').css('background-color');
if (color == 'rgb(0, 0, 255)' || color == 'blue') // =='blue' <- IE hack
    alert("it's blue!\nColor detected: " + color);

The above sample will only work for the first element in the JQuery selector, therefore you should use the .each JQuery method to iterate through the whole selector if you have more than one element in it.

Please note that the .css JQuery method returns a RGB combination in most browsers (except IE), so testing it against a string such as blue will not suffice for non-IE browsers. You can test that in the JQuery API site and my fiddle.

And here’s with the proper .each iteration:

$('.box').each(function(i){
    var color = $(this).css('background-color');
    if (color == 'rgb(0, 0, 255)' || color == 'blue') // =='blue' <- IE hack
        alert("div " + i + " is blue!\nColor detected: " + color);
});​

JSFiddle

---

Edit: over a year and half later, I feel like this answer deserves an update.

For most use cases, I’d recommend using a CSS class:

.blue {
    color: blue;
}

This allows the usage of the addClass(), removeClass(), toggleClass() methods for manipulation, as well as hasClass() for checking:

if ( $('#myElem').hasClass('blue') ) {
    //it has the .blue class!
}

This works seamlessly in all browsers without needing hacks, and as a plus you get better separation of concerns — that is, if your styling ever changes, say .blue { color: lightblue; }, the JS will keep working without modifications.

Note: of course, this approach is not suitable for a couple rare use cases where the color value is dynamically generated (e.g. using Math.random() or picking a color value off a canvas pixel), in these rare use cases you can still use the first solution in this answer.