Ticket #4518 (closed bug: invalid)
fadeIn cannot handle "display: inline !important;"
| Reported by: | acoder | Owned by: | acoder |
|---|---|---|---|
| Priority: | major | Milestone: | 1.4 |
| Component: | effects | Version: | 1.3.2 |
| Keywords: | Cc: | ||
| Blocking: | Blocked by: |
Description
jQuery's fadeIn works fine with "display: inline;" but fails with "display: inline !important;". Instead of fadeIn the element appears immediately.
Change History
comment:2 Changed 4 years ago by dmethvin
Do you mean that you have an element with an inline style of "display: inline !important"? Can you attach a test case and/or explain further?
comment:3 Changed 4 years ago by acoder
Ok, I think this is what happened:
I have an "<a>" element that was assigned a "display: inline;" via class declaration. I set a "display: none;" via the style declaration so that fadeIn would work. The style declaration overrode the external css declaration. When I set the external css to "display: inline !important;", however, the css file overrode the style declaration. I guess the problem is that fadeIn apparently only works if the element is set to "display: none;" (correct me if I'm wrong). Right now I am using fadeTo as a workaround.
What I am hoping for is a way of using fadeIn with "display: inline;" elements.
Please follow the bug reporting guidlines and use jsFiddle when providing test cases and demonstrations instead of pasting the code in the ticket.