Delete a pattern present in file 2 from file 1 if found in file 1.

Shell Programming and Scripting
1

I have two files

File1
====
1|2000-00-00|2010-02-02||
2| 00:00:00|2012-02-24||
3|2000-00-00|2011-02-02||

File2
====
2000-00-00
 00:00:00

I want the delete the patterns which are found in file 2 from file 1,

Expected output:

File1
====
1||2010-02-02||
2||2012-02-24||
3||2011-02-02||

I did something like

while read line
  do
     sed 's/'$line'//g' File1 > File1_tmp
     mv File1_tmp File1
  done < File2

Not sure if this is good approach. Can someone help me with some better solution?

2
  1. This would only remove that part of string from file1 which matches with file2. It won't remove the whole line from file1.
  2. You may do this:
while read line
do
    grep $line file1 >> file1_tmp
done < file2
mv file1_tmp file1
  1. Or you may also do this:
grep -fv file2 file1
3

I just wanted to remove the data in file1 which matches the pattern in file2. Not the whole line.:o

4

Hi machomaddy,

One way using perl:

$ cat file1
1|2000-00-00|2010-02-02||
2| 00:00:00|2012-02-24||
3|2000-00-00|2011-02-02||
$ cat file2
2000-00-00
 00:00:00
$ cat script.pl
use warnings;
use strict;

die qq[Usage: perl $0 <file-1> <file-2>\n] unless @ARGV == 2;

open my $fh1, qq[<], shift @ARGV or die qq[Cannot open input file\n];
open my $fh2, qq[<], shift @ARGV or die qq[Cannot open input file\n];

my (@patterns, $pattern, $re);

while ( <$fh2> ) {
        next if m/\A\s*\Z/;
        s/\A\s+/\\s*/;
        s/\s+\Z/\\s*/;
        push @patterns, $_;
}

$pattern = join qq[|], @patterns;
$re = qr/($pattern)/;

while ( <$fh1> ) {
        do { print, next } if m/\A\s*\Z/;
        chomp;
        s/$re//go;
        printf qq[%s\n], $_;
}
$ perl script.pl file1 file2
1||2010-02-02||
2||2012-02-24||
3||2011-02-02||

Regards,
Birei

5

See if this works:

awk -F\| 'NF==1{A[$1];next}$2 in A{$2=x}1' OFS=\| file2 file1
6

try this

awk -F, 'NR==FNR{a[$0]=$0;next} {for(i in a) gsub(i,""); print} ' file_with_pattern file