extract date from filename

star7 has asked for the wisdom of the Perl Monks concerning the following question:

I wrote following commands in my script: (testing)

my $file = "textfile.020701_00:08";
my $time = substr($file, 9,12);
$time =~ s/(\d{6})(\d{1})(\d{2})(\d{1})(\d{2})/"$1$3$5.monday"/;

print "TIME: $time"
[download]

What i want is the following result: 0207010008.monday on my display Can you specify bugs in this code. I need your help.star7

Comment on extract date from filename - need help Download Code

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Re: extract date from filename - need help by Lachesis (Friar) on Aug 12, 2003 at 16:33 UTC
Your problem is that your regex effectively boils down to /\d{12}/ ignoring the substitution and variable capture, so it will only match a 12 digit number. You actually need to match the _ and : characters. What you want would be something like. `my $file = "textfile.020701_00:08"; my $time = substr($file, 9,12); $time =~ s/(\d{6})_(\d{2}):(\d{2})/"$1$2$3.monday"/; print "TIME: $time";` [download]	[reply] [d/l]
Re: extract date from filename - need help by Not_a_Number (Prior) on Aug 12, 2003 at 17:34 UTC
Why complicate matters? `my $file = "textfile.020701_00:08"; $file =~ s/\D//g; my $time = $file . '.monday'; print $time;` [download] dave	[reply] [d/l]
Re: extract date from filename - need help by sweetblood (Prior) on Aug 12, 2003 at 16:39 UTC
Well the "\d" in your expression means a digit so when you say s/(\d{6})(\b{1}) you are saying find 6 digits and assign them to $1 then find 1 digit and assign it to $2. Your string contains 6 digit followed by an "_" so you expression will fail. if you wish to go this way you would want s/^(\d{6})_(\d{2}):(\d{2})/$1$2$3.monday/ This should give you what you want. You'll notice that I added the "^" to the beginning of the expression to anchor my search to the leftmost position.	[reply]