Hi Lisaraël
thanks for your help and sorry for late reply - I was down with a flu :-(
the odd idea was a great idea and it works perfectly :clap: It would also
select the first td, but this is not a problem as its not input ready.
thanks again
GGerri
Lisaraël wrote:
$('#PG1_L03-table
Karl Rick
thanks a lot for your help and sorry for late reply - I was down with a flu
:-(
Karl's version with the each loop works fine. Lisaraël's solution is a bit
simpler and works as col 1 is not input ready and all 'odd' td should get a
binding.
Thanks though for teaching me another way
On Nov 12, 2008, at 9:05 AM, Rik Lomas wrote:
This is completely untested, but this is a bit simpler:
$('#PG1_L03-table').find('td:eq(3), td:eq(5), td:eq(7),
td:eq(9)').find('input').bind..
That won't work, because it'll only select the 4th, 6th, 8th, and 10th
TD.
2008/11/12
Following on from Karl, how about:
$('#PG1_L03-table td:nth-child(2n+3)').slice(0, 4).find('input')
2008/11/12 Lisaraël [EMAIL PROTECTED]:
$('#PG1_L03-table td:nth-child(odd) input') ? (untested too)
--
Rik Lomas
http://rikrikrik.com
Ah yeah, damn you :nth-child for being one-indexed
2008/11/12 Karl Swedberg [EMAIL PROTECTED]:
On Nov 12, 2008, at 9:05 AM, Rik Lomas wrote:
This is completely untested, but this is a bit simpler:
$('#PG1_L03-table').find('td:eq(3), td:eq(5), td:eq(7),
On Nov 12, 2008, at 11:13 AM, Rik Lomas wrote:
Following on from Karl, how about:
$('#PG1_L03-table td:nth-child(2n+3)').slice(0, 4).find('input')
That won't work either, unless the table has only one row. Would have
to do something like this instead:
$('#PG1_L03-table
This is completely untested, but this is a bit simpler:
$('#PG1_L03-table').find('td:eq(3), td:eq(5), td:eq(7),
td:eq(9)').find('input').bind..
Rik
2008/11/12 ggerri [EMAIL PROTECTED]:
Hi Gurus
have the following selector and wonder how this could be simplified?
:confused:
Haha, I'll give up trying to code on the fly and actually test my JS next time
2008/11/12 Karl Swedberg [EMAIL PROTECTED]:
On Nov 12, 2008, at 11:13 AM, Rik Lomas wrote:
Following on from Karl, how about:
$('#PG1_L03-table td:nth-child(2n+3)').slice(0, 4).find('input')
That won't work
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