从文件中读取一行，如果模式匹配，则在 perl 中将其删除

Question

我有一个文件 place.txt，内容如下：

I want to go Rome. Will Go.
I want to go Rome. Will Not Go.
I want to go Rome. Will Not Go.
I want to go Rome. Will Go.
I want to go India. Will Not Go.
I want to go India. Will Not Go.
I want to go Rome. Will Go.

我想读取这个文件，并将行与“我想去罗马”模式匹配。并在 perl 中省略与此文件中的模式匹配的那些行。

我的示例代码是：

$file = new IO::File;
$file->open("<jobs.txt") or die "Cannot open jobs.txt";

while(my $line = $file->getline){
    next if $line =~ m{/I want to go Rome/};
    print $line;
}

close $file;

注意：我的文件会很大。我们可以使用 sed 或 awk 吗？

Answer 1

就这么简单

perl -ine'print unless /I want to go Rome/'

如果你喜欢脚本

use strict;
use warnings;
use autodie;

use constant FILENAME => 'jobs.txt';

open my $in, '<', FILENAME;

while (<$in>) {
    if ( $. == 1 ) {    # you need to do this after read first line
                        # $in will keep i-node with the original file
        open my $out, '>', FILENAME;
        select $out;
    }
    print unless /I want to go Rome/;
}

Answer 2

grep -v '我想去罗马'jobs.txt

写起来还是简单多了。

Answer 3

awk '$0~/I want to go Rome/' jobs.txt

Answer 4

试试这个：

use strict;
use warnings;
use File::Copy;

open my $file, "<", "jobs.txt" or die "Cannot open jobs.txt : $!";
open my $outtemp, ">", "out.temp" or die "Unable to open file : $!\n"; # create a temporary file to write the lines which doesn't match pattern.

while(my $line = <$file>)
{
    next if $line =~ m/I want to go Rome/; # ignore lines which matches pattern
    print $outtemp $line; # write the rest lines in temp file.
}
close $file;
close $outtemp;
move("out.temp", "jobs.txt"); # move temp file into original file.