perl帮助根据时间输入解析日志文件

我是新手。我有一个日志文件，需要解析出“备份成功”和任何“错误：”条目。我尝试使用unix cat解析日志文件，并将其传输到grep。我得到了我想要的信息，但是我想在perl中尝试一下，并且还可以选择传递一个日期参数，并根据我需要的日期给我行

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

日志文件输出示例：（备份成功）

日志文件输出示例：（错误：）

**我想要一个文本和/或电子邮件与此信息。像这样，但是可以选择我需要的日期

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，请为我提供一些perl代码和/或开始使用的想法。我将非常感谢你的帮助。谢谢。

这里有一个简单的脚本。要扫描的文件名和目标日期是硬编码的。匹配项打印到标准输出

顺便说一句，这段代码完全没有经过测试。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;        
}

使用严格；
使用警告；
我的$logpath='./bar/log'；
我的美元目标=‘2010年6月9日’；
打开我的$fh，“
#”/usr/bin/perl
#用法示例：2010年6月9日
严格使用；
使用警告；
我的（$mon，$day，$year）=（$ARGV[0]、$ARGV[1]、$ARGV[2]）；
打开（FH，“<$ARGV[3]”）或死亡“无法打开日志文件$ARGV[3]：$！\n”；
while（我的$line=）{
如果（$line=~/.$mon$day\d{2}:\d{2}:\d{2}$year:.*（错误：|备份成功）/）{
打印$行；
}
}
完美！这很有效。正是我需要的。谢谢你的帮助！
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;        
}

#!/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;
}
}