如何使用bash中的参数从文件中提取行[重复]

Question

我有一个包含所有信息的日志文件。

2017-09-21 02:01:11,130 [http-nio-8080-exec-5] INFO   - Inbound Message
...
...
2017-09-21 09:01:11,130 [http-nio-8080-exec-5] INFO   - Inbound Message
----------------------------
ID: 6044
Address: http://localhost/serveur/Service?wsdl
Encoding: UTF-8
Http-Method: POST
...
2017-09-21 12:01:11,130 [http-nio-8080-exec-5] INFO   - Inbound Message
...

我只想提取两个日期之间的信息。例如，如果我想要 2017-09-21 09.. 和 2017-09-21 12 之间的信息，我会有这样的信息：

./script.sh 2017-09-21 09 2017-09-21 12
# $1 = 2017-09-21
# $2 = 09
# $3 = 2017-09-21
# $4 = 12
# $1 & 2 are date and hour to begin extract
# $3 & 4 are date and hour to finish extract

我的脚本是这样的。我没有shell编程知识，但我尝试过这样做。不幸的是它不起作用

#!/bin/bash

beginLine = "$1 $2"
finishLine = "$3 $4"

while read fileLog.log
do
    if($fileLog.log grep beginLine)
         echo $fileLog.log >> extractedlog.log
    fi
    if($fileLog.log grep finishLine)
         < extractedLog.log;
    fi
done
exit 0;

谁能告诉我问题出在哪里？

Answer 1

$ awk '/2017-09-21 09/{a=1};a;/2017-09-21 12/{exit}' input
2017-09-21 09:01:11,130 [http-nio-8080-exec-5] INFO   - Inbound Message
----------------------------
ID: 6044
Address: http://localhost/serveur/Service?wsdl
Encoding: UTF-8
Http-Method: POST
...
2017-09-21 12:01:11,130 [http-nio-8080-exec-5] INFO   - Inbound Message

找到更多替代品here

Answer 2

短sed方法（匹配一系列模式/.../,/.../）：

sed -n '/2017-09-21 09/,/2017-09-21 12/p' logfile

---------

awk（甚至更短）也可以做到这一点：

awk '/2017-09-21 09/,/2017-09-21 12/' logfile