CODEWITHSUNDEEP
regexperl
misaka
misaka

Reputation: 29

Perl Regular Expression Question

I wrote a Perl program which reads text from text file and prints it out.

I want to print out a line which has specific format.

For example, there are some lines like this:

information:
Ahmad.prn:592118:2001:7:5:/Essay
Ashford.rtf:903615:2001:6:28:/usr/Essay
Barger.doc:243200:2001:7:4:/home/dir
end of Information.

I want to read only these three lines:

Ahmad.prn:592118:2001:7:5:/Essay
Ashford.rtf:903615:2001:6:28:/usr/Essay
Barger.doc:243200:2001:7:4:/home/dir

I think that the meaning of the fields is:

Ahmad.prn <- file name
592118 <- size of file
2001:7:5 <- created date
/Essay <- path of file

My code is this:

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

open (my $infh, "<", $file)||die "cant open";

while(my $line = <$infh>) {
    chomp ($line);
    if ($line =~ /(what regular expression do I have to put in here?)/) {
        print "$line";
    }
}

close ($infh);

Upvotes: 2

Views: 727

Answers (4)

Rajesh
Rajesh

Reputation: 11

$line =~ ([a-zA-Z.]+):(\d+):(\d+):(\d+):(\d+):([\/A-Za-z]+)

$name = $1; #Ahmad.prn
$id = $2; #592118
$year = $3; #2001
$dir = $6; #/Essay

Note: loop through it for multiple names

Upvotes: 0

thematta
thematta

Reputation: 1

#!/usr/bin/perl
use strict;

my $inputText = qq{
Ahmad.prn:592118:2001:7:5:/Essay
Ashford.rtf:903615:2001:6:28:/usr/Essay
Barger.doc:243200:2001:7:4:/home/dir
end of Information.
};

my @input = split /\n/, $inputText;
my $i = 0;
while ($input[$i] !~ /^end of Information.$/) {
    if ($input[$i] !~ /:/) {
        $i++;
        next;
    }
    my ($fileName, $fileSize, $year, $month, $day, $filePath) = split /:/, $input[$i];
    print "$fileName\t $fileSize\t $month/$day/$year\t $filePath\n";
    $i++;
}

Upvotes: 0

SAN
SAN

Reputation: 2247

Since you have this format for Ahmad.prn:592118:2001:7:5:/Essay

Ahmad.prn <- file name
592118 <- size of file
2001:7:5 <- created date
/Essay <- path of file

you can use this regular expression

/^\s*(\S+):(\d+):(\d+:\d+:\d+):(\S+)\s*$/

With this you will have file name in $1, Size of the file in $2, Date of creation in $3, Path to the file in $4

I added optional spaces in the start and end of the line, if you want to allow optional spaces after or before : you can add \s*

Upvotes: 0

YOU
YOU

Reputation: 123791

If lines you need always ends with /Essay, you may use following regex

/:\/Essay$/

Edit 1: looks there is middle parts are only numbers, you may match this way.

/:\d+:\d+:\d+:\d+:/

Upvotes: 3

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