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Re: regex help
by sauoq (Abbot) on Sep 27, 2002 at 20:44 UTC
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Your question is poorly stated. If you always want "/ASP/scripts/homesites/" then you could just set a variable to that. You wouldn't need a regex. So, I suspect that you want to match a certain class of strings.
That could be a class of strings that always consists of the 2nd through 3rd directories in the path component of a URI. That could be a class of strings that consists of "/ASP" followed by two more directories. etc. etc. etc.
What do you really want to match? Can you provide more examples?
-sauoq
"My two cents aren't worth a dime.";
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Re: regex help
by hiseldl (Priest) on Sep 27, 2002 at 20:41 UTC
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Re: regex help
by Cody Pendant (Prior) on Sep 28, 2002 at 08:54 UTC
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As an aside, try not to use $a, because it's a special variable used for sorting (comparing $a against $b).
--
($_='jjjuuusssttt annootthheer
pppeeerrrlll haaaccckkeer')=~y/a-z//s;print;
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Re: regex help
by fglock (Vicar) on Sep 27, 2002 at 20:40 UTC
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s|/ASP/scripts/homesites/||;
Is this what you mean? | [reply]
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I cnat use that cause I have other ones that are like "/browse/CGI/scripts/" that are totally different
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Re: regex help
by andrew (Acolyte) on Sep 27, 2002 at 21:24 UTC
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Alright
$a =~ s|/browse(/.*/)|$1|i; $a = $1;
That seems to work :), but what if the url conatins '\' instead of '/' like how could I check for both. | [reply]
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[\\\/]
or:
[\/\\]
so you get:
$a =~ s|[\/\\]browse([\/\\].*[\/\\])|$1|i;
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Re: regex help
by andrew (Acolyte) on Sep 27, 2002 at 20:47 UTC
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I just wanna match everything after browse and everything before the last "/" | [reply] |
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while ( <DATA> )
{
if ( /browse(\/.*)\// ) { #if the current $_ matches
#browse followed by a
# / followed by anything
#but having to be followed by
# another /
#then put whatever was found
#between the parens in $line
my $line = $1;
print "$line\n";
}
}
__DATA__
http://www.url.com/browse/ASP/scripts/homesites/
/browse/ASP/scripts/homesites/
/browse/ASP/scripts/homesites/index.htm
I don't know if you noticed but in your original post browse was misspelled.
-Enlil | [reply]
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while ( <DATA> )
{
if ( /browse(\/.*\/)/ ) { #if the current $_ matches
#browse followed by a
# / followed by anything
#but having to be followed by
# another /
#then put whatever was found
#between the parens in $line
my $line = $1;
print "$line\n";
}
}
__DATA__
http://www.url.com/browse/ASP/scripts/homesites/
/browse/ASP/scripts/homesites/
/browse/ASP/scripts/homesites/index.htm
I meant to put the last \/ inside the parens as you wanted the final / included in the match at the end of the returned string. My mistake. -Enlil | [reply]
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my ($match) = $a =~ m!(/browse.*/)!;
Use that and $match will contain the slash prior to "browse" and everything following up to and including the final slash.
Be forewarned, that will break if you give it a string ending in a directory that you want to match but which doesn't have a slash. The string 'http://foo.bar.com/some/path/browse/ASP/scripts', for example, would cause $match to be set to '/browse/ASP/'. If that's alright, then great. If not, they you have to find another way of distinguising a file and a directory. (Perhaps all of your files contain a dot and none of your directories do.)
-sauoq
"My two cents aren't worth a dime.";
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Well, everything after browse and everything before the last "/"
can be also like this
$a =~ s|/browse(.*/)|$1|i;
Comment: Ooops! Too late! It's like hiseldl
Hopes
perl -le '$_=$,=q,\,@4O,,s,^$,$\,,s,s,^,b9,s,$_^=q,$\^-]!,,print'
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