Question

Perl新手在这里。我有一个日志文件，我需要解析“备份成功”和任何“错误：”条目。我尝试使用unix cat解析日志文件并将其传递给grep。我得到了我想要的信息，但我想在perl中尝试这个，并且还可以选择传递日期参数，并根据我需要的日期给我一行。

日志文件输出示例:(备份成功）

Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: flush-logs-time=00:00:00
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-time=06:14:23
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-status=Backup succeeded
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: Backup succeeded

日志文件输出示例:(错误:)

Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: mysql-zrm appears to be already running for this backupset
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: If you are sure mysql-zrm is not running, please remove the file /etc/mysql-zrm/rip1.mail.mad/.mysql-zrm.pid and restart mysql-zrm

**我希望收到包含此信息的文字和/或电子邮件。像这样，但可以选择传递我需要的日期。

Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: mysql-zrm appears to be already running for this backupset
Wed Jun 09 05:00:03 2010: rip1.mil.mad:backup:ERROR: If you are sure mysql-zrm is not running, please remove the file /etc/mysql-zrm/rip1.mail.mad/.mysql-zrm.pid and restart mysql-zrm
Wed Jun 09 06:14:25 2010: db2.cal.mil.mad:backup:INFO: backup-status=Backup succeeded

如果您愿意，请提供一些perl代码和/或想法，以便开始使用。我很感激你的帮助。谢谢。

Answer 1

#!/usr/bin/perl

# usage example: <this script> Jun 09 2010 <logfile>

use strict;
use warnings;

my ($mon,$day,$year) = ($ARGV[0],$ARGV[1],$ARGV[2]);

open(FH,"< $ARGV[3]") or die "can't open log file $ARGV[3]: $!\n";

while (my $line = <FH>) {
    if ($line =~ /.* $mon $day \d{2}:\d{2}:\d{2} $year:.*(ERROR:|Backup succeeded)/) {
    print $line;
}
}

Answer 2

这是一个简单的脚本。要扫描的文件名和目标日期是硬编码的。匹配打印到STDOUT。

顺便说一句，这段代码完全未经测试。我在浏览器的文本框中输入了它。

use strict;
use warnings;

my $logpath = './bar/log';
my $target = 'Jun 09 2010';

open my $fh, '<', $logpath or die "Error opening $logpath $!\n";

while (my $line = <$fh> ) { 

    next unless date_match( $target, $line );

    next unless my $result = got_error($line) // got_backup($line);    

    print $result;
}

sub got_backup {
     my $line = shift;

     return unless $line =~ /backup-status=Backup succeeded/;

     return $line;
}

sub got_error {
     my $line = shift;

     return unless $line =~ /:ERROR:/;

     return $line;
}


# Take a line and a target date.  Compare the date derived from the line to
# the target, and returns true if they match.
# Also always returns true if target is not defined

sub date_match {
    my $target = shift;
    my $line = shift;

    return 1 unless defined $target;  # Always true if target is undefined.

    # Where did that god-awful date format come from?  Yech.
    my $date = extract_date($line);

    return $date eq $target;
}

# Simple extract of date using split and join with extra variables
# to make it newbie friendly.
# IMO, it would be a good idea to switch to using DateTime objects and 
#    DateTime::Format::Strptime

sub extract_date {
    my $line = shift;

    my @parts = split /:/, $line;
    my $date = join ':' @parts[0..2];
    @parts = split /\s+/, $date;

    $date = @parts[1,2,4]; 

    return $date;        
}

您可以使用Getopt::Long获取文件名和目标日期。
使用更强大的日期/时间解析和比较方案是个好主意。 DateTime和朋友是非常好的，强大的日期操作模块。检查一下。
如果您处理大量数据并且需要提高效率，则可以避免以多种方式在任何地方复制$line。
为了将来参考，如果您发布一些代码，您会得到更好的回复

perl帮助根据时间输入解析日志文件

2 个答案: