I don't think you're really understanding the selector syntax.
The 'id' is unique (or is supposed to be) and starts with a '#' in a
selector, and
the css syntax uses a name and value pair to set the attribute. So:
$("#myimage).css('display", "inline"); // or "block"
or try this alternative:
$("#myimage).show();
or possibly:
$("#myimage).toggle()
The last one displays when hidden and hides if already visible. I
assume that 'myimg' is not something you really intended to use for
the selector.
On Dec 9, 7:16 am, JQueryProgrammer <[EMAIL PROTECTED]> wrote:
> <image id="myimage" myimgid="myimageid" src="baloon.jpg" />
>
> I want to check the style display for this image. I am doing as:
>
> $("image[myimgid='myimageid']").css("display");
>
> but its coming undefined. I cannot check it with id as my id is
> getting runtime generated. Please help.
