note 25415 added to function.preg-replace

Regarding the last note on replacement strings containing numbers -- which in turn get seen as part of the replacement pattern when substituting like so: \\1$replacement. The date example fix may work, but I'm doing many replacements on a database so I wanted to get it right with one line if possible. My line of text used some pipe (|) delimiters to split up some numbers. Rather than exploding them, updating the changes and finally rebuilding the line, I used something like this: $text = '1|2|3|4|5|6|7|8|9|10'; $pattern = '/(.*)((\|.*){2})((\|.*){3})/U'; $replacement = '|100|100|'; echo preg_replace($pattern,"\\1\\2$replacement",$text); The key for my needs was making the replacement text/numbers surrounded by the pipe delimiters. This way the first character is not treated as part of the n'th match (i.e. \\1$replacement becomes \\1|100|100| instead of \\1100|100). By using the [U]ngreedy modifier and making the first matchable sub-pattern a (.*), the second sub-pattern actually matches {n+1} items between the pipe delimiters. Since it's actually 2 pattern matches put together, the replacement substitution is \\1\\2 instead of only \\1 ... assuming you want to keep the first portion of the original text as you found it. From that point in the replacement pattern (after \\1\\2) your replacement string is inserted. Your sub-pattern match count ({3} in the example for 3 pipes) has to pair up with the number of pipes/elements you're sticking in as a replacement. This match in the pattern makes for the 3rd sub-pattern match or \\3 BTW. The remainder of the original string is returned as-is since it doesn't match anything after the last or 3rd sub-pattern (not matched, not replaced). Adding another (.*) sub-pattern to the end of the pattern (creating a \\4) would zap the remainder of the string, ending it after your inserted text. This worked for my case but might not if you need the replacement right at the beginning of the original string. With a little tweaking and TLC, it should work for most cases where delimiters are used. If anyone knows or can suggest a better way to work around the "\\1$numbers" problem, don't hesitate to holler. Getting a pattern to match as you expected is always a great way to spend the weekend ;) Enjoy! -- http://www.php.net/manual/en/function.preg-replace.php http://master.php.net/manage/user-notes.php?action=edit+25415 http://master.php.net/manage/user-notes.php?action=delete+25415 http://master.php.net/manage/user-notes.php?action=reject+25415