Question

我看了，但没有找到答案。我想读取一个日志文件并打印出“：”之后的所有内容，但有些日志之前有空格。我想在开头只匹配一个没有空格的那个。

_thisnot: this one has space
thisyes: this one has not space at the beginning.

我想对文件中的每一行都这样做。

Answer 1

怎么样：

#!/usr/bin/perl 
use strict;
use warnings;
use 5.010;

my %result;
while(<DATA>) {
    chomp;
    next if /^\s/;
    if (/^([^:]*):\s*(.*)$/) {
        $result{$1} = $2;
    }
}

__DATA__
 thisnot: this one has space
thisyes: this one has not space at the beginning.

Answer 2

或者你可以使用像衬里一样的衬里：

perl -ne 's/^\S.*?:\s*// && print' file.log

Answer 3

# Assuming you opened log filehandle for reading...
foreach my $line (<$filehandle>) {
    # You can chomp($line) if you don't want newlines at the end
    next if $line =~ /^ /; # Skip stuff that starts with a space
              # Use  /^\s/ to skip ALL whitespace (tabs etc)
    my ($after_first_colon) = ($line =~ /^[^:]*:(.+)/);
    print $after_first_colon if $after_first_colon; # False if no colon. 

}

Perl - 正则表达式读取日志文件

3 个答案: