It doesn't work because browsers return a CSSStyleDeclaration object,
or something similar, you can try it with firebug: $('body')[0].styleBut I have just released a "plugin" that enables you to solve this very easily. I put some other stuff into it too that you probably don't need but that can be easily removed. Here it is: http://plugins.jquery.com/project/EnhancedAttributeSelectors and this how it can be used: $("div[~background-color=firebrick]"); The standard way would be to make a pseudo selector for this but somehow I like this approach more. If you don't, then have a look at the plugins page, I'm sure someone has made a lighter pseudo selector for this. On Jan 29, 8:47 pm, kevotheclone <kevothecl...@gmail.com> wrote: > Thanks for the reply Shane! > > Yes looping through the parent is an option that I was hoping to > avoid. Rigth now my Timeline only displays 30 projects (each one > composed of 2 <div> elements), but SIMILE Timeline generates many more > <div> elements within the parent containing <div>; 380 to be exact. > So I'll have to loop through 380 to find the a few <div> elements that > have the appropriate background color. Since I don't generate any of > the Timeline elements I'm kind of at SIMILE mercy (and I don't wat to > alter their code). I guess performance wise me examining 380 <div> > elements isn't any different than waht JQuery would have to do, but I > was hoping for a nice elegant JQuery one liner. > > It does seem like $("div[style*= firebrick]"); should work. > > Find all <div> elements whose "style" attribute contains "firebrick". > > Can anyone explain the logic why this doesn't work? > Is this a bug or a documented exception that I failed to find/read? > > Thanks again to Shane and the rest of the JQuery community!
