Re: Regular Exprections

2003-11-25 Thread Hacksaw
I'm betting you don't get to dictate that the string has no newlines in it. -- A principle is an instruction in qualitative endeavour. http://www.hacksaw.org -- http://www.privatecircus.com -- KB1FVD -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED

RE: Regular Exprections

2003-11-25 Thread EUROSPACE SZARINDAR
Marcos, Try this : my $str = 'aaa--%%b%bb%%--ccc--%%ddd%%--'; $str =~ s/(-{2}\%{2})(.*?)(\%{2}-{2})/$1$3/g; print "Modified str = ".$str."\n"; you will get Modified str = aaa----ccc---- Is it what you want ?? Michel -Message d'origine- De: [EMAIL PROTECTED] [mailto:[EMA

RE: Regular Exprections

2003-11-25 Thread Perry, Alan
Hacksaw wrote: > > $str =~ s/--\%\%.*?\%\%--/--\%\%\%\%--/sg; You don't need all those backslashes. This works as well (and IMHO is easier to read): $str =~ s/--%%.*?%%--/----/sg; Alan -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]

Re: Regular Exprections

2003-11-25 Thread Hacksaw
t.pl: #!/usr/bin/perl -w my $str = 'aaa --%% b%b %%-- ccc --%%ddd%%--'; $str =~ s/--\%\%.*?\%\%--/--\%\%\%\%--/sg; print $str, "\n"; -- hacksaw > perl t.pl aaa ---- ccc ---- -- All the magic is in the matching operator. The ? after the .* makes it match the