在bash中简单的正则表达式解析

Question

我想解析一个包含类似这些行的日志文件(log.txt):

2010-10-19 07:56:14 URL:http://www.website.com/page.php?ID=26 [13676] -> "www.website.com/page.php?ID=26" [1]
2010-10-19 07:56:14 URL:http://www.website.com/page.php?ID=44 [14152] -> "www.website.com/page.php?ID=44" [1]
2010-10-19 07:56:14 URL:http://www.website.com/page.php?ID=13 [13681] -> "www.website.com/page.php?ID=13" [1]
2010-10-19 07:56:14 ERROR:Something bad happened
2010-10-19 07:56:14 ERROR:Something really bad happened
2010-10-19 07:56:15 URL:http://www.website.com/page.php?ID=14 [12627] -> "www.website.com/page.php?ID=14" [1]
2010-10-19 07:56:14 ERROR:Page not found
2010-10-19 07:56:15 URL:http://www.website.com/page.php?ID=29 [13694] -> "www.website.com/page.php?ID=29" [1]

你可能已经猜到了:

1)我需要从每一行中提取这部分:

2010-10-19 07:56:15 URL:http://www.website.com/page.php?ID=29 [13694] -> "www.website.com/page.php?ID=29" [1]
------------------------^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

2)这部分转到另一个文件(log.html),如下所示:

<a href="http://www.website.com/page.php?ID=29">http://www.website.com/page.php?ID=29</a>

我需要通过bash脚本执行此操作,该脚本将在*nix平台上运行.我不知道shell编程,所以详细的脚本将非常感激,指向bash编程参考的指针会做.

Answer 1

这是一个bash解决方案

#!/bin/bash
exec 4<"log.txt"
while read -r line<&4
do
  case "$line" in
    *URL:* )
      url="${line#*URL:}"
      url=${url%% [*}
      echo "<a href=\"${url}\">${url}</a>"
  esac
done
exec 4<&-