On Jun 13, 5:21 am, [EMAIL PROTECTED] (James) wrote: > Thanks all, I have something working > > > $data =~ s/(.*\n)(?=\1)//g; > > Can anyone explain the (?=\1) bit? I get the search replace.
Which part do you not understand? The (?=) or the \1 or both? (?= ) is a "positive lookahead assertion". It "peeks" into the pattern match to determine if the next thing matches its contents, but it doesn not actually match those contents. It doesn't move the internal position pointer along, and whatever is in the (?= ) is not part of the actual match so will not be replaced. \1 within a pattern match means exactly what $1 will mean when the pattern match is finished. That is, it's whatever was matched by the first capturing parentheses in this pattern match. In this case, that's .*\n. So this pattern is searching for 0 or more of any character, followed by the newline, and then checks to see if the next thing after that is exactly what was matched again. If so, the entire *MATCH* is replaced with nothing. Since the second instance was of the .*\n was not actually matched, just looked for, it does not get replaced. For more information, perldoc perlretut perldoc perlre perldoc perlreref Paul Lalli -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
