James, Actually the interpreter will evaluate the '\1'. '1' . '\2' into one string before passing it to preg_replace So it will yield up as I originally wrote it, and it won't work.
Elias "Jason Wong" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > On Friday 16 August 2002 19:49, lallous wrote: > > <? > > $fn = 'test.gif'; > > > > echo preg_replace('/(.+?)(\..+?)/', '\1a\2', $fn); > > > > ?> > > > > This script will output 'testa.gif' > > > > now how can i make it produce 'test1.gif' ? > > > > if i replace the: '\1a\2' with '\11\2' it will understand to replace with > > occurence number 11 ! > > How can i escape the 2nd '1' so it is considered as a string an not another > > number. > > Try: > > echo preg_replace('/(.+?)(\..+?)/', '\1'. '1' . '\2', $fn); > > -- > Jason Wong -> Gremlins Associates -> www.gremlins.com.hk > Open Source Software Systems Integrators > * Web Design & Hosting * Internet & Intranet Applications Development * > > /* > IBM's original motto: > Cogito ergo vendo; vendo ergo sum. > */ > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
