输入：
--------------- 1st block ----------------
info1:AAAAA
random info
random info
random info
info0:"one string"
random info
random info
[...]
info2:"2nd string"
info3:"3rd string"
random info
random info
info4:"4th string"
--------------- 2nd block ----------------
info1:BBBBB
random info
random info
random info
info0:"one string"
random info
random info
[...]
info2:"2nd string"
info3:"3rd string"
random info
random info
info4:"4th string"
--------------- 3rd block ----------------
[...]
注意：如果info0是我搜索的模式，我想获取info0，info1，info2，info3，info4
我可以执行grep info0 -B 4（以获取info1），但是我不能执行-A x来获取其余信息（info2,3,4），因为行的数量在一个块与另一个块之间可以不同。
重要：
-the output should be grouped by info0
-info0, info2,info3,info4 is always the same, but info1 is always unique (actually info1 is a MAC address).
-the number of blocks is not limited (there can be 1, 2 , 3, 4,...blocks).
-subsidiary question: within the same command, do the same thing but with info0 carrying different string. For instance: 2 blocks with the same info0, then 3 blocks with another info0 and so on and so forth...
- I put ----- #block----- in the input to facilitate lecture to the reader, but these lines do not exist in the actual input.
想要的输出：
info0:"one string"
info1:AAAAA
info1:BBBBB
info2:"2nd string"
info3:"3rd string"
info4:"4th string"
到目前为止我做了什么
output | grep info0 -B 4 -A 23 | grep '\(info1\|info0\|info1\|info2\|info3\|info4\)' >> /tmp/info0
23->估计一个块的最大行数
非常感谢你们！
答案 0 :(得分：-1)
我找到了一个坚固但可行的解决方案。假设我搜索info0，输入为/ tmp / input：
echo info0;grep info0 -B5 /tmp/input | grep info1;awk '/info0/{flag=1} flag; /info4/{flag=0}' /tmp/input | grep '\(info2\|info3\)' | sort -u