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Regex Question

summergs

Anyone know the Regex for extracting the last digit from the following:

$DCRFilename = DCRName_09-15-2004_3

I am trying to extract the last digit (may be a digit between 0 and 9). DCRName can be of differing lengths.
my $test = ($DCRFilename =~ m^*\_*\_(.*?); Doesnt cut it

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Comments

Dwayne

How about

my ($test) = ($DCRFilename =~ m/(\d)$/);

--

Current project: TS 5.5.2/6.1 W2K

nipper

You could always chop:

$test = $DCRFilename;
$lastchar = chop($test);

Chop removes the last character from the variable & returns it.

That will work IF the last character is always the one you want.

Andy

Migrateduser

Dwayne's answer is the best approach. Regex-wise, what you have would have almost worked, but you munged the syntax a little. For one thing, the * quanitifier after the ^ positional anchor is meaningless. Also, you don't need to escape underbars, they aren't a metacharacter. This is probably what you intended.

my ($test) = ($DCRFilename =~ m|^.*_*_(.*?)| );

However, it's bad, bad, bad, because the regex engine would have to do a lot of trackbacks to finally get the match. Dwayne's answer should execute with something on the order of a quarter of the number of required operations. Given that this is the second or third regex question you've asked, I would strongly encourage you to pick up Jeff Friedl's Mastering Regular Expressions book. It will change your life.

summergs

Thanks!

Dwayne

This is probably what you intended.

my ($test) = ($DCRFilename =~ m|^.*_*_(.*?)| );

Actually, I think this is more what was intended. (Easy mistake to make)

my ($test) = ($DCRFilename =~ m|^.*_.*_(.*?)| );

But, again, if you know that you always want the last character in all cases, it's much better to use the earlier, simpler regex, unless you also want to ensure that you only match for filenames that follow the pattern given.

And, "change your life"? It's a good book, but that might be just a tad strong. ;-)

--

Current project: TS 5.5.2/6.1 W2K

Edited by Dwayne on 09/16/04 07:48 AM (server time).

summergs

Ok. As fate (in this case Management) would have it, I need to capture the other part of the File Name.
i.e. If the File name is DCR_File_Name_09-14-2004.html, then I need to return DCR_File_Name.... if the file name is ABCDEFG_12_09-14-2004.html, then I need to return ABCDEFG_12... if the file name is ABCDEFG_09-14-2004.html then return ABCDEFG.

Any Ideas?

Thanks in advance!

Adam Stoller

If there's no path attached (or if you want to keep the path) - try:

my ($val) = m|^(.*)_.*\.html|;

if you want to remove any preceding path and just get the value - try:

my ($val) = $_ =~ m|^.*/(.*)_.*\.html$|;

- I believe they will provide you the value (in $val) that you're looking for - based on the examples you've just given.

--fish
Senior Consultant, Quotient Inc.
http://www.quotient-inc.com

summergs

Thanks Fish... If the file had no extention would the code be?
my ($val) = m|^(.*)_.*$|;

Adam Stoller

That would probably work - the easiest way to do this is create a list of the kind of filenames you expect to encounter, write a regex you think will do the job and create a little test script the processes the list with the regex and prints out the results.

--fish
Senior Consultant, Quotient Inc.
http://www.quotient-inc.com

summergs

Unfortunately I can not predetermine the filenames. Only that the filenames will be in a pattern of filename_mm-dd-yyyy. The filename itself may contain 0 zero or more "_"

I could not get a return from m|^(.*)_.*|;
Any other ideas?

Adam Stoller

The parentheses around the variable:

my ($val) = $string =~ m|(pat)tern|;

is important - without it you'll only get something like '1' as a return value.

If you know that it's the last part that you can count on you could also do something like:

if ($string =~ m!^(.*)\d{2,2}-\d{2,2}-\d{4,4}(|\.html)$!){
    $val = $1;
}

--fish
Senior Consultant, Quotient Inc.
http://www.quotient-inc.com

summergs

That was it. Thanks! (Also thanks for the info on the parens as I did not know that)

Migrateduser

That's not something peculiar to regexes, either. It's a basic perl concept regarding the difference between scalar and list contexts. Several functions or operators behave a bit differently depending on whether you are treating their return value as a scalar or a list. Enclosing the lvalue of an assignment in parens forces list context.

Here's a pretty decent article that I found via google that will give you the basics of what you need to know about context:

http://www.itworld.com/nl/perl/03012001/