[jQuery] Re: Image.css("display") not working
Now I understand you. 'id' was in so many places I thought you were
confused about how to use them. After rereading, I see you refer to
as the tag, did you really mean that or did you use in
your html? is not the proper HTML tag name, it is
On Dec 9, 11:24 pm, JQueryProgrammer <[EMAIL PROTECTED]> wrote:
> Well I think you did not understand my quest correctly. As I said I
> cannot use the id for selector as my id is getting runtime generated.
> So i am using a custom attribute "myimgid" to select the element. I
> tried to use event.preventDefault too. but still its firing the
> original attached event. Any help would be appreciated.
>
> On Dec 9, 6:24 pm, donb <[EMAIL PROTECTED]> wrote:
>
>
>
> > 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:
>
> > >
>
> > > 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.- Hide quoted text -
>
> - Show quoted text -
[jQuery] Re: Image.css("display") not working
Well I think you did not understand my quest correctly. As I said I
cannot use the id for selector as my id is getting runtime generated.
So i am using a custom attribute "myimgid" to select the element. I
tried to use event.preventDefault too. but still its firing the
original attached event. Any help would be appreciated.
On Dec 9, 6:24 pm, donb <[EMAIL PROTECTED]> wrote:
> 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:
>
> >
>
> > 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.
[jQuery] Re: Image.css("display") not working
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:
>
>
> 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.
[jQuery] Re: Image.css("display") not working
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:
>
>
> 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.
