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modifying a program to open all files in a directory

oldrob
Novice

Mar 13, 2013, 5:17 PM

Post #1 of 3 (164 views)

modifying a program to open all files in a directory	Can't Post

Hello Perl Gurus,

I have written a script to take input files, copy specific data and output a combined file (see code below). Id like to make it so that I just have to specify a directory on the command line so that the script will automatically work on all the contained files - any help much appreciated!

Code

 
#a program to parse flow data 
#!/usr/bin/perl 
use warnings; 
my @buffer; 
my @data; 
 
foreach (@ARGV) { 
	$filename = $_; #sets the filename variable 
	open (my $fh, '<',$filename); 
	$lineno = 1; 
	while (<$fh>) { 
		$lineno++; 
		if ($lineno == 3) { 
			chomp(@buffer = <$fh>); #copy the entire line to @buffer 
			unshift(@buffer, "$filename \t"); #add the filename before the string 
			push (@buffer, "\n"); #add a newline after the string 
			push (@data, @buffer); 
		} 
		 
	} 
} 
 
open OUTPUT, '>', "output.txt"; 
print OUTPUT "@data";

Chris Charley
User

Mar 13, 2013, 5:54 PM

Post #2 of 3 (160 views)

Re: [oldrob] modifying a program to open all files in a directory [In reply to]	Can't Post

Code

chomp(@buffer = <$fh>); #copy the entire line to @buffer

This doesn't copy line 3 to @buffer. It copies lines 3 to eof to the buffer.

Update: Ok, I guess thats what you want the program to do - sorry.

You can use the glob function to get files like below.

Code

my $dir = shift @ARGV; 
my @cvs_files = glob "$dir/*.csv";

You would just need to specify your directory on the command line immediately after the program, like perl my_program.pl dir_name

(This post was edited by Chris Charley on Mar 14, 2013, 8:59 AM)

oldrob
Novice

Mar 14, 2013, 3:01 PM

Post #3 of 3 (138 views)

Re: [Chris Charley] modifying a program to open all files in a directory [In reply to]	Can't Post

Many thanks Chris, this did the trick nicely!

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