On 2010.07.09 21:57, Uri Guttman wrote: >>>>>> "SB" == Steve Bertrand <st...@ipv6canada.com> writes: > > SB> On 2010.07.09 21:40, Uri Guttman wrote: > > >> what input? i see none there. but what you want is either -p or -n. both > >> are among the most useful options for perl oneliners. look them up in > >> perlrun and pick which one you want. > > SB> | perl -p -e 's/.*\s+//' > > you need to test that one liner all by itself by feeding it some of > those lines. the regex is very wrong as it will delete the entire line > and therefore print nothing. .* matches the whole line before the > newline and \s+ matches the trailing newline. if you added the -l option > too, it would strip the newline off before regex is run and likely will > do what you want. but a better regex is needed. i gotta run so i can't > help with that now.
Thanks for the help all, here's what I've come up with: perl -pe 's/(.*?)\s+//' ...match the least amount possible before one or more spaces, and replace everything up to the last space with nothing. It now also works without -l, which removes the line ending before processing, and then adds it back afterwards (per perldoc perlrun). Apparently, the reason -l made my original regex work is because there was no newline, so the \s+ stopped in the tab in between the output, and didn't run until the end of the string, as Uri mentioned in his description. Thanks everyone. Steve -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
