capturing names

Sara has asked for the wisdom of the Perl Monks concerning the following question:

hello , I got this data in my file

nbsssbs/cn/bld/cn.bld:cn_commonsw_libs.bld
nbsssbs/cn/bld/cn_commonsw_libs.bld:#   File: cn_commonsw_libs.bld
[download]

I would like to capture the .bld , I want

$hold1 = cn.bld;
$hold2 = cn_commonsw_libs.bld;
[download]

the problem I am having is , I am trying to avoid the .bld followed by :# in the second line .. , I just need to pare line like the first one

cn.bld:cn_commonsw_libs.bld
[download]

thanks for helps and ideas

2002-07-09 Edit by Corion : Added formatting

Comment on capturing names Select or Download Code

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Re: capturing names by hiseldl (Priest) on Jul 09, 2002 at 15:01 UTC
It looks like you wanted to comment out the last part of line 2 with #. So you could try removing the comment before you parse the line: `$line2 =~ s/#.*$//;` [download] this should give you: `$hold1 = cn_commonsw_libs.bld; $hold2 = ""; # empty` [download] -- .dave.	[reply] [d/l] [select]
Re: capturing names by Rich36 (Chaplain) on Jul 09, 2002 at 14:12 UTC
Take a look at split - this function allows you to split strings based on whatever character you specify and returns a list of elements derived from the split up string. In this instance, you could detect for the existance of the '#', then split on it and get the first element of the array that split returns - which would be the first part of the line that you want to capture. «Rich36»	[reply]
Re: capturing names by aseidas (Beadle) on Jul 09, 2002 at 15:14 UTC
After Dave's regex (and suggestion about the comment) try `for ($data){ if (m#/([^/]*\.bld)$#) { $count=$count+1; $hold[$count]=$1; print "$hold[$count]\n"; } }` [download] Aseidas	[reply] [d/l]