I don't think you're not really understanding the selector syntax. The 'id' is unique (or is supposed to be) and starts with a '#', 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.