reformat text in perl

Question

I have a file of 1000 lines, each line in the format

filename dd/mm/yyyy hh:mm:ss

I want to convert it to read

filename mmddhhmm.ss

been attempting to do this in perl and awk - no success - would appreciate any help

thanks

Answer 1

You can do a simple regular expression replacement if the format is really fixed:

s|(..)/(..)/.... (..):(..):(..)$|$2$1$3$4.$5|

I used | as a separator so that I do not need to escape the slashes.

You can use this with Perl on the shell in place:

perl -pi -e 's|(..)/(..)/.... (..):(..):(..)$|$2$1$3$4.$5|' file

(Look up the option descriptions with man perlrun).

Answer 2

Another somehow ugly approach: foreach line of code ($str here) you get from the file do something like this:

my $str = 'filename 26/12/2010 21:09:12';

my @arr1 = split(' ',$str);
my @arr2 = split('/',$arr1[1]);
my @arr3 = split(':',$arr1[2]);

my $day = $arr2[0]; 
my $month = $arr2[1]; 
my $year = $arr2[2];

my $hours = $arr3[0]; 
my $minutes = $arr3[1]; 
my $seconds = $arr3[2];

print $arr1[0].' '.$month.$day.$year.$hours.$minutes.'.'.$seconds;

Answer 3

Pipe your file to a perl script with:

while( my line = <> ){
    if ( $line =~ /(\S+)\s+\(d{2})\/(\d{2})/\d{4}\s+(\d{2}):(\d{2}):(\d{2})/ ) {
        print $1 . " " . $3 . $2 . $4 . $5 . '.' . $6;
    }
}

Redirect the output however you want. This says match line to: (non-whitespace>=1)whitespace>=1(2digits)/(2digits)/4digits whitepsace>=1(2digits):(2digits):(2digits)

Capture groups are in () numbered 1 to 6 left to right.

Answer 4

Using sed:

sed -r 's|/[0-9]{4} ||; s|/||; s/://; s/:/./' file.txt

• delete the year /yyyy
• delete the remaining slash
• delete the first colon
• change the remaining colon to a dot

Using awk:

awk '{split($2,d,"/"); split($3,t,":"); print $1, d[1] d[2] t[1] t[2] "." t[3]}'