Sorry, I need the number (marked with x): forum-list-x
and not the string before the number. On Jun 15, 3:55 pm, "Ariel Flesler" <[EMAIL PROTECTED]> wrote: > var n = +"forum-list-1".slice(-1); ;-) > > -- > Ariel Fleslerhttp://flesler.blogspot.com > > On 6/15/08, Klaus Hartl <[EMAIL PROTECTED]> wrote: > > > > > > > And in case you want to shorten the parseInt: > > > var n = +"forum-list-1".split('-').pop(); > > > --Klaus > > > On 15 Jun., 14:24, Klaus Hartl <[EMAIL PROTECTED]> wrote: > > > var n = parseInt(/\d/.exec("forum-list-1")[0]); > > > > Or even easier using split (if you can rely on the format of the > > > string) - pretty much what Ariel already showed: > > > > var n = parseInt("forum-list-1".split('-').pop()); > > > > --Klaus > > > > On 15 Jun., 13:18, yo2lux <[EMAIL PROTECTED]> wrote: > > > > > Thanks and which is the best way to obtain the last number of string, > > > > for example I have: > > > > > forum-list-1 > > > > forum-list-2 > > > > forum-list-3 > > > > > I need the numbers: 1, 2 or 3. > > > > > On Jun 12, 6:44 pm, Karl Swedberg <[EMAIL PROTECTED]> wrote: > > > > > > And that's why Ariel is da man! Nice one. > > > > > > --Karl > > > > > ____________ > > > > > Karl Swedbergwww.englishrules.comwww.learningjquery.com > > > > > > On Jun 12, 2008, at 7:09 AM, Ariel Flesler wrote: > > > > > > > Ooooooor.... > > > > > > > var fileName = $('img').attr('src').split('/').pop(); > > > > > > > Cheers :) > > > > > > > -- > > > > > > Ariel Flesler > > > > > >http://flesler.blogspot.com/ > > > > > > > On 11 jun, 22:29, Karl Swedberg <[EMAIL PROTECTED]> wrote: > > > > > >> Hi there, > > > > > > >> You could do it like this: > > > > > > >> var path = $('img').attr('src'); > > > > > >> var fileName = path.slice(path.lastIndexOf('/')+1); > > > > > > >> --Karl > > > > > >> ____________ > > > > > >> Karl Swedbergwww.englishrules.comwww.learningjquery.com > > > > > > >> On Jun 11, 2008, at 7:36 PM, yo2lux wrote: > > > > > > >>> path variable store the following: > > /themes/mytheme/images/logo.gif: > > > > > > >>> var path = $('img').attr('src'); > > > > > > >>> is possible to obtain the image name without path ? I need > > logo.gif > > > > > >>> I need a Javascript or jQuery function to solve this? > > > > > > >>> Thanks!- Ocultar texto de la cita - > > > > > > >> - Mostrar texto de la cita - > > -- > Ariel Fleslerhttp://flesler.blogspot.com
