与多列进行文件比较

Question

I am doing a directory cleanup to check for files that are not being used in our testing environment. I have a list of all the file names which are sorted alphabetically in a text file and another file I want to compare against.

Here is how the first file is setup:

test1.pl
test2.pl
test3.pl

It is a simple, one script name per line text file of all the scripts in the directory I want to clean up based on the other file below.

The file I want to compare against is a tab file which lists a script that each server runs as a test and there are obviously many duplicates. I want to strip out the testing script names from this file and compare spit it out to another file, use uniq and sort so that I can diff this file with the above to see which testing scripts are not being used.

The file is setup as such:

server: : test1.pl test2.pl test3.pl test4.sh test5.sh

There are some lines with less and some with more. My first impulse was to make a Perl script to split the line and push the values in an list if they are not there but that seems wholly inefficient. I am not to experienced in awk but I figured there is more than one way to do it. Any other ideas to compare these files?

Answer 1

这通过awk将文件名重新排列为第二个文件中的每行一个，然后将输出与第一个文件区diff 。

diff file1 <(awk '{ for (i=3; i<=NF; i++) print $i }' file2 | sort -u)

Answer 2

A Perl solution that makes a %needed hash of the files being used by the servers and then checks against the file containing all the file names.

#!/usr/bin/perl
use strict;
use warnings;
use Inline::Files;

my %needed;
while (<SERVTEST>) {
    chomp;
    my (undef, @files) = split /\t/;
    @needed{ @files } = (1) x @files;
}

while (<TESTFILES>) {
    chomp;
    if (not $needed{$_}) {
        print "Not needed: $_\n";   
    }
}

__TESTFILES__
test1.pl
test2.pl
test3.pl
test4.pl
test5.pl
__SERVTEST__
server1::   test1.pl    test3.pl
server2::   test2.pl    test3.pl
__END__
*** prints

C:\Old_Data\perlp>perl t7.pl
Not needed: test4.pl
Not needed: test5.pl

Answer 3

Quick and dirty script to do the job. If it sounds good, use open to read the files with proper error checking.

use strict;
use warnings;
my @server_lines = `cat server_file`;chomp(@server_lines);
my @test_file_lines = `cat test_file_lines`;chomp(@test_file_lines);
foreach my $server_line (@server_lines){
   $server_line =~ s!server: : !!is;
   my @files_to_check = split(/\s+/is, $server_line);
   foreach my $file_to_check (@files_to_check){
      my @found = grep { /$file_to_check/ } @test_file_lines;
      if (scalar(@found)==0){
        print "$file_to_check is not found in $server_line\n";
      }
   }

}

Answer 4

If I understand your need correctly you have a file with a list of tests (testfiles.txt):

test1.pl
test2.pl 
test3.pl
test4.pl
test5.pl

And a file with a list of servers, with files they all test (serverlist.txt):

server1:        :       test1.pl        test3.pl
server2:        :       test2.pl        test3.pl

(Where I have assumed all spaces as tabs).

If you convert the second file into a list of tested files, you can then compare this using diff to your original file.

cut -d: -f3 serverlist.txt | sed -e 's/^\t//g' | tr '\t' '\n' | sort -u > tested_files.txt

The cut removes the server name and ':', the sed removes the leading tab left behind, tr then converts the remaining tabs into newlines, then we do a unique sort to sort and remove duplicates. This is output to tested_files.txt .

Then all you do is diff testfiles.txt tested_files.txt .

Answer 5

It's hard to tell since you didn't post the expected output but is this what you're looking for?

$ cat file1
test1.pl
test2.pl
test3.pl
$
$ cat file2
server: : test1.pl test2.pl test3.pl test4.sh test5.sh
$
$ gawk -v RS='[[:space:]]+' 'NR==FNR{f[$0]++;next} FNR>2 && !f[$0]' file1 file2
test4.sh
test5.sh