How to extract the filenames in an array

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Re: How to extract the filenames in an array by GrandFather (Saint) on Sep 17, 2008 at 04:27 UTC
Most likely you want to use something like File::Basename to manipulate file names and paths. Nice to see you are using warnings. You will find that strict (`use strict;`) helps a lot too. Perl reduces RSI - it saves typing	[reply] [d/l]
Re: How to extract the filenames in an array by lamp (Chaplain) on Sep 17, 2008 at 05:24 UTC
If you want to find all the filename of a particular directory, you can use 'opendir' and 'readdir' function. For eg. `opendir(DIR, "/tmp") \|\| die "can't opendir $!"; while (my $file = readdir(DIR)) { next unless (-f "/tmp/$file"); print $file."\n"; } closedir DIR;` [download] If you want check a particular file name, you can use the '-e' option. For eg. `my $fname = "/tmp/test.txt"; print $fname if(-e $fname);` [download]	[reply] [d/l] [select]
Re^2: How to extract the filenames in an array by parv (Parson) on Sep 17, 2008 at 05:30 UTC
The `-e` function checks for existence of a given path (among other functions which check for other things, of course).	[reply] [d/l]
Re: How to extract the filenames in an array by CountZero (Bishop) on Sep 17, 2008 at 06:08 UTC
`use strict; my $dirpath = '/usr/C'; opendir(my $dir, $dirpath) or die "Error when using opendir $!"; my @namearr = grep {-f "$dirpath/$_"} readdir($dir); # get only files +and skip directories` [download] It is usually more efficient to use Perl's own internal functions rather than shelling out to an external command or program. CountZero A program should be light and agile, its subroutines connected like a string of pearls. The spirit and intent of the program should be retained throughout. There should be neither too little or too much, neither needless loops nor useless variables, neither lack of structure nor overwhelming rigidity." - The Tao of Programming, 4.1 - Geoffrey James	[reply] [d/l]
Re: How to extract the filenames in an array by jwkrahn (Abbot) on Sep 17, 2008 at 05:08 UTC
`@name1=$namearr[0];` is the same as saying `$name1[0]=$namearr[0];` so there is nothing in `$name1[2]` to print out.	[reply] [d/l] [select]
Re: How to extract the filenames in an array by repellent (Priest) on Sep 17, 2008 at 06:44 UTC
Welcome to PerlMonks! Let's say we have the path: `/login/user/temp/file.txt` Imagine splitting the path names into 5 containers. This is how we would do it: `#!/usr/bin/perl -l use warnings; use strict; use File::Spec::Functions qw(splitdir); my $path = '/login/user/temp/file.txt'; my @names = splitdir $path; print $names[0]; # prints: "" print $names[1]; # prints: "login" print $names[2]; # prints: "user" print $names[3]; # prints: "temp" print $names[4]; # prints: "file.txt"` [download] The first array element is an empty string because nothing comes before the first directory separator of `$path`.	[reply] [d/l] [select]
Re: How to extract the filenames in an array by Grey Fox (Chaplain) on Sep 17, 2008 at 14:36 UTC
Howdy; It's also good to ask all of your related questions in one post. I see you asked a similar question in How do i take the first 4 characters of any file name in a specified directory? -- Grey Fox "We are grey. We stand between the darkness and the light" B5	[reply]