CODEWITHSUNDEEP
regexperlunix
gkns
gkns

Reputation: 720

Variable inside grep in Perl

I have to search for a pattern inside a variable in Perl. The pattern also needs to be inside a variable. Here is what I have:

opendir(DIR,"reports/");
$poolname = $poolname."_";
@FILES= grep {/^$poolname\.[0-9]*.csv/} readdir(DIR);
@sorted= reverse sort @FILES;

The pattern I want to match is poolname_[0-9]* (I'm trying to get the latest report for a pool here. [0-9]* is the unix timestamp of the file)

But the above regex is not working as expected. $sorted[0] doesn't have the required filename. May I know what is wrong with the above code?

Upvotes: 0

Views: 16175

Answers (1)

Jonathan Leffler
Jonathan Leffler

Reputation: 753575

Assuming that somewhere before the snippet in the question, there is an assignment equivalent to:

my $poolname = "poolname";

then you say you are searching for:

poolname_[0-9]*          # Presumably, poolname_[0-9]*.csv in fact

but your regex is searching for:

poolname_\.[0-9]*.csv    # Probably should have a backslash before the .csv

The patterns you seek will not be matched by your regex; remove the \. to get the result you require.

opendir(DIR,"reports/") or die "$!";
@FILES  = grep { /^${poolname}_[0-9]*\.csv/ } readdir(DIR);
@sorted = reverse sort @FILES;
closedir(DIR);

Given a directory reports containing files:

x_1332827070.csv
x_1333366051.csv
x_1333A66051.csv
y_1332827070.csv

this script:

#!/usr/bin/env perl
use strict;
use warnings;

my $poolname = "x";
opendir(DIR,"reports/") or die "$!";
my @FILES  = grep { /^${poolname}_[0-9]*\.csv/ } readdir(DIR);
my @sorted = reverse sort @FILES;
closedir(DIR);

print "$_\n" for @sorted;

produces the output:

x_1333366051.csv
x_1332827070.csv

If that is not what you're after, your comments and your question are misleading.

Upvotes: 5

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