Reputation: 6185
I have this slidetoggle and I want the style of the open toggle to be different then the closed ones.
By default all the faqtopics1
are set to border-radius: 5px; background-color: #f2ecec;
when the div faqtext
associated opens.
When the toggle opens, I want the style of faqtopics1
to be set to the "OnClick Style"
border-radius: 5px 5px 0 0;
background-color: #dedcdc;
I found out about the .css() Method and could somehow make something up (line 2 and 3):
$(".faqtopics1").click(function(event) {
$("div.faqtopics1").css({"border-radius":"5px", "background-color":"#f2ecec"});
$(this).css({"border-radius":"5px 5px 0 0", "background-color":"#dedcdc"});
$("div.faqtext").stop(true).slideUp(400);
$(this).next("div.faqtext").stop(true).slideToggle();
});
But it's not a total success as even when I re-click on a toggle to close it, the OnClick style remains. Is there a better way to make what I want ?
Also I want to apply the same principal even if I click on faqtopics2, faqtopics3 or faqtopics4 div. (cf the jsfiddle).
You can find my codes (css + query) on this jsfiddle
Thanks a lot for your help!
Upvotes: 0
Views: 82
Reputation: 7714
Something much easier:
Define your two states in CSS:
faqtopics1 {
border-radius: 5px;
background-color: #f2ecec;
}
.onclickstyle {
border-radius: 5px 5px 0 0;
background-color: #dedcdc;
}
Then in JS you just have to toogle the class:
$("div.faqtopics1").toggleClass("onclickstyle");
This means you have a clear separation between the exact style (in the css), and the dynamic toogle (in the javascript).
Upvotes: 2
Reputation: 873
It may be easier to use addClass.
$this.addClass('active');
Then in your css
.faqtopics.active{border-radius:5px 5px 0 0; background-color:#dedcdc;}
You can give all of your "FAQ topics" a shared class .faqtopics
and then unique id's #faqtopic1 #faqtopic2
if you need to style them a bit differently.
Upvotes: 1