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Find and replace last line in a file

Hi i am having multiple files with different date at last line which s at 2nd position

is it possible to achieve to change the 2nd position date to one single date
for eg:

Code:

file1

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,12-Jun,t



file2

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,10-Jun,t



file3

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,16-Dec,t

i need to replace in all the files last line like below without affecting any other data in each of the files

Code:

file1

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,10-Aug,t



file2

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,10-Aug,t



file3

Aaa,bb,cc,dd

Xxx,yy,d,12

Dodd,10-Aug,t

Show us your attempt and where you are stuck? If not sure where to start I would suggest sed as this should be rather simple.

This what i have tried

sed -e '$s/10-Jun/10-Aug/' file its working but since i am having multiple files and also multiple dates to replace with 10-Aug i am struck on it

If each group is always 4 lines followed by an empty line you could just do “p” three times, followed by “s/,.*,/,10-Aug/” and another “p”.

Or, if the line you want to change always starts with “Dodd”, you can just use that in your “s” command.

Hi,

its not that always its 4 lines and start with same "Dodd". I have just given a sample example but it contains N number of line and starts with any character data

If that's the only date in the files and the formatting is consistent across all files, you could try replacing it as follows:

Code:

sed -n 's/[0-9][0-9]*-[A-Z][a-z][a-z]/10-Aug/' files

It will match: [1 or 2 digits] - [capital letter][lower case letter][lower case letter] as in 12-Aug

hmmm ... I was thinking if it is always 3 fields separated by commas just grab the lot:

Code:

sed '$s/,.*,/,10-Aug,/' files

Yes, do what grail said. I didn't notice that those are separate files and you want to change the last line of each of those files. Furthermore, if you want to make the change in place and you have a GNU sed, you can use “-i” switch.

This Perl script will replace the last line of any text file with whatever you want it to be. It has options to write the file back to the original and/or the default just prints to standard out.

Code:

#!/usr/bin/perl

# replaceLastLine.pl

# sydney 12/28/2013

# This script is provided for rohit_shinez and has no warranty. 

# It is in response to http://www.linuxquestions.org/questions/programming-9/find-and-replace-last-line-in-a-file-4175489359/



use strict;

use warnings;



# Command Line Arguments

my $file = shift( @ARGV );

my $replace = shift( @ARGV );

my $write = "0";

my $help = "0";

my $suppress = "0";



for ( @ARGV ) {

        if (( $_ eq "-w" ) || ( $_ eq "--write" )) {

                $write = "1";

        } 

        if (( $_ eq "-h" ) || ( $_ eq "--help" )) {

                $help = "1";

        }

        if (( $_ eq "-s" ) || ( $_ eq "--suppress" )) {

                $suppress = "1";

        }

}



# Help Section and Usage

if (

        ( not defined $file ) ||

        ( not defined $replace ) ||

        ( $file eq "-h" ) || 

        ( $file eq "--help" ) ||

        ( $replace eq "-h" ) || 

        ( $replace eq "--help" ) ||

        ( $help eq "1" ) 

) {

        print "Usage: replaceLastLine.pl [FILE] \"replacement text\" [OPTIONS] \n  -h, --help                print help \n  -w,--write                Replace the original file.\n  -s,--suppress                Suppress standard output (do not print to screen)\n"; 

        exit;

}



# Open File

my $fh;

my @lines;

open ( $fh, '<' , $file ) or die;

# Collect the Lines

while ( <$fh> ) {

        push ( @lines,$_ );

}

close $fh;

# Remove the last line

pop( @lines );

# Add new last line

push ( @lines , $replace . "\n" );

# Print to Standard Out

if ( $suppress eq "0" ) { 

        print @lines;

}

# Print New File

if ( $write eq "1" ) {

        open ( $fh, '>' , $file ) or die;

        print $fh @lines;

        close $fh;

}

if you have Ruby

Code:

contents  = File.open("file").readlines      # read whole file

last_line = contents[-1].split(",")          # split last line

last_line[1] = "new date"                    # change date

contents[-1] = last_line.join(",")            # join back

File.open( "file", "w+" ).write( contents )  # update

@Sydney; you can replace the loop file-into-array

Code:

while ( <$fh> ) {

        push ( @lines,$_ );

}



#Use this one-line 'slurp' method instead

@lines = <$fh>

Quote:

#Use this one-line 'slurp' method instead
@lines = <$fh>

Nice Thanks