> -----Original Message----- > From: rszeus [mailto:rszeus@xxxxxxxxx] > Sent: 22 July 2009 19:23 > To: 'Jim Lucas' > Cc: 'Kyle Smith'; 'Eddie Drapkin'; ash@xxxxxxxxxxxxxxxxxxxx; php- > general@xxxxxxxxxxxxx > Subject: RE: Replace in a string with regex > > No, sory, my bad typing. It's not the problem.. > I have the same on both caxses, only chnage the variable $id > $file = "screen/temp/7a45gfdi6icpan1jtb1j99o925_1_mini.jpg" > $id = '70'; > echo preg_replace('#(screen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, > $file); > Get: 0 > > file = "screen/temp/7a45gfdi6icpan1jtb1j99o925_1_mini.jpg"; > $id = test; > echo preg_replace('#(screen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, > $file); > > Get: screen/test > > What is wrong with having na integer on the var ? Well, the problem here is that when you concatenate $id containing 70 on to '$1', you effectively end up with '$170' -- which the manual page at http://php.net/preg-replace makes clear is ambiguous, but is likely to be treated as $17 followed by a zero, rather than $1 followed by 70. Since $17 doesn't exist (as you don't have that many capturing subpatterns in your pattern!), it is taken to be the null string -- leaving just the left over 0 as the result of the replacement. QED. The workaround for this is also given on the page referenced above, which is to make your replacement be '${1}'.$id. Incidentally, I don't really see the need for the $1, or the equivalent parentheses in the pattern -- since a fixed string is involved, why not just use it directly in both places? Like this: preg_replace('#screen/temp/(.+?)_1(.+?\.jpg)#', 'screen/'.$id, $file); which completely sidesteps the problem. Cheers! Mike -- Mike Ford, Electronic Information Developer, Libraries and Learning Innovation, Leeds Metropolitan University, C507, Civic Quarter Campus, Woodhouse Lane, LEEDS, LS1 3HE, United Kingdom Email: m.ford@xxxxxxxxxxxxxx Tel: +44 113 812 4730 To view the terms under which this email is distributed, please go to http://disclaimer.leedsmet.ac.uk/email.htm
