linux - 找到模式时将行转为列

Question

这是我要处理的示例文件。我想使用关键字“描述”作为某种 RS，但不知道该怎么做，而且它也不一致。

背景:我正在处理一个日志文件，其中第一行包含日期/时间戳 (APR12)，第二行有关于日志的描述。此描述适用于少数日志和 missig。

001 APR12 aaa bbb
Description: This is a test file.
002 APR12 aaa bbb
Description: This is another test file.
003 APR12 aaa XXX
004 APR12 aaa bbb
Description: This is another,after skipping one. 

期望的输出:

001 APR12 aaa bbb Description: This is a test file.
002 APR12 aaa bbb Description: This is another test file.
003 APR12 aaa XXX
004 APR12 aaa bbb Description: This is another,after skipping one.

Answer 1

$ awk '{printf "%s%s", (/^[0-9]/?rs:FS), $0; rs=RS} END{print ""}' file
001 APR12 aaa bbb Description: This is a test file.
002 APR12 aaa bbb Description: This is another test file.
003 APR12 aaa XXX
004 APR12 aaa bbb Description: This is another,after skipping one.