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What is the proper way to do fallbacks with .css()? This would most commonly be used for font fallbacks but I am using it for cursor images.

Here's what I got that isn't working:

$("#dragable").css('cursor','url(images/grabbing.gif), -moz-grabbing, auto');

**UPDATE: can someone tell me the valid CSS for starters?

What I have:

cursor: url(images/grabbing.gif), -moz-grabbing, auto;

... doesn't work.**

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  • For font fallbacks you should use the standard CSS syntax, eg jQueryObj.css({ fontFamily: '"Primary Font", "Fallback Font", serif' });. Other properties will be another matter and I'll leave that up to the hopefully forthcoming answers. Commented Nov 23, 2010 at 22:36

2 Answers 2

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Using dynamic inline styles not always a good idea. Consider dynamic classes. Define somewhere in css specific class for draggable element:

.cur-draggable { cursor: url(images/grabbing.gif), -moz-grabbing, auto; }

And try to add this class, instead of style itself;

$("#dragable").addClass('cur-draggable');
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1 Comment

... this makes life easier. Good idea.
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Not sure why that breaks it, but the values aren't being written to the DOM.

Though, the css() function is really just a shortcut to editing the "style" attribute inline on the element, so you could do this to get the same result:

$("#dragable").attr('style','cursor: url(images/grabbing.gif), -moz-grabbing, auto');

2 Comments

Note that this will overwrite other style properties.
True, but you could easily bypass that by first retrieving the contents of the style attribute and then appending the string. Something like $("#dragable").attr('style',$(this).attr('style') + 'cursor: url(images/grabbing.gif), -moz-grabbing, auto'); -- not sure that exact syntax will work, but the principle is there.

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