opening file using wild card

Question posted by: Looking2Learn (Newbie) on July 3rd, 2008 01:29 PM

Hey,
I'm trying to open a txt file, but I won't know the first three characters of the filename. For example, I want to open a "user.txt" file, and I know it will have three digits in front of it, like "123_user.txt", but they could be any three digits. Is there a way to introduce a wild card into the open command to specify this?

Code: ( text )

#!usr/bin/perl
open (FILE, "***_user.txt") or die "Can't find file!" ;
print "Found file!" ;
close (FILE) ;

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nithinpes

Expert

253 Posts

July 3rd, 2008
02:56 PM
#2

Re: opening file using wild card

Wildcard cannot be used with open function. Are you trying to open any file/all files in the current directory that matches your criteria?
I am assuming that you want to search for files that match the criteria and open them. In this case, you can parse through the directory and fetch all files that match the criteria and open them for reading as below:

Code: ( text )

#!usr/bin/perl
use strict;
my @files;
opendir(DIR,".") or die "opening directory failed:$!"; # '.' for pwd.Use dir path if required.
while(my $filename=readdir(DIR)){
push @files,$filename if($filename=~/^\d\d\d_user\.txt$/);
}
closedir(DIR);
foreach my $file (@files) {
print "found $file\n";
open (FILE, "$file") or die "Can't find $file!" ;
while(<FILE>) {
### read
}
close (FILE) ;
}

Last edited by nithinpes : July 3rd, 2008 at 02:58 PM. Reason: added comment

nithinpes

Expert

253 Posts

July 3rd, 2008
03:00 PM
#3

Re: opening file using wild card

In case you want to open any single file that match the criteria assign $filename to $file when a match is found and quit the while(my $filename=readdir(DIR)) {} loop.

Looking2Learn

Newbie

5 Posts

July 3rd, 2008
03:15 PM
#4

Re: opening file using wild card

Great, thanks!

And yes, there will only be one matching file. So, I'll use:

Code: ( text )

while(my $filename=readdir(DIR)){
my file = $filename if($filename=~/^\d\d\d_user\.txt$/);
}

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