jQuery Selector equivalent of javascript

Question

I am trying to get the jquery equvaliant of this javascript

var id = $(this).parent().parent().parent().attr("id");
document.getElementById(id).getElementsByClassName("addcomment")[0].style.display = 'block';

but its not working

$('#+id+' '.addcomment').css('display','block');

Any suggestions ?

If you're using a code editor, then these sort of errors are caught/identified very easily by the syntax highlighting... — nickf, May 18 '12 at 10:20
possible duplicate of [How to use javascript variables in jquery selectors](http://stackoverflow.com/questions/5891840/how-to-use-javascript-variables-in-jquery-selectors) — Felix Kling, May 18 '12 at 10:20
Might be worth posting some html as you are getting some very presumptuous answers — musefan, May 18 '12 at 10:21

Fabrizio Calderan · Accepted Answer · 2012-05-18T10:31:47.687

4

$('#' + id + '.addcomment').css('display','block');

as a sidenote in the page you should have only one element with that id, so

$('#' + id ).css('display','block');

should works too (of course only if classname it's not necessary to target it, since this is a different selector)

edited May 18 '12 at 10:31

answered May 18 '12 at 10:18

Fabrizio Calderan

Your second example is incorrect as it will apply block to the `#id` element, not the `#id .addcomment` element. – Rory McCrossan May 18 '12 at 10:20
1

Or `$(this).parent().parent().parent().find('.addcomment').css('display','block');` – robertc May 18 '12 at 10:20
@RoryMcCrossan of course they're different selectors. it's just to suggest another solution if classname it's not necessary to target that element – Fabrizio Calderan May 18 '12 at 10:22

score 1 · Answer 2 · answered May 18 '12 at 10:19

1

$('.addcomment', '#' + id).css('display','block');

or just simple

$('#' + id).css('display','block');

answered May 18 '12 at 10:19

thecodeparadox

score 1 · Answer 3 · edited May 18 '12 at 11:02

1

I think you need to get the child elements based on their class. So try this.

$('#' + id ).find('.addcomment').show();

edited May 18 '12 at 11:02

Mathew Thompson

answered May 18 '12 at 10:57

indika

3 Answers3