在AWK或SED中找到带有正则表达式的字符串 [英] Found string with regex in AWK or SED

问题描述

例如，我想在File中找到以下字符串9Stest1.test2D9.然后我要剪切前2个字符，最后2个字符，最后在befor和after之后打印文本.在两条单独的线中.

I want to find for example the following string 9Stest1.test2D9 in File. Then i want to cut the first 2 charechter and last 2 charechter and finally print the text befor and after . in two seperate Line.

示例文字:

7U8vTest(#G-HLjYM6QqJj1j"7MFx$^Qd
.f@alU|A#Z<inCWV6a=L?o`A5vIod"%Mm+YW1RM@,L;aN
r^n<&)}[??!VcVIV**9Stest1.test2D9**94EN~yK,$lU=9?UT.[
e`)G:FS.nGz%?@~k!20aLJ^PU-[@}0W\ !8x
cujOmEK"1;!cI134lu%0-A +/t!VIf?8uT`!
aC1QAQY>4RE$46iVjAE^eo5yR|
1?/T?<H5,%G~[|9I/c&8MY$O]%,UYQe{!{Bm[rRC[
aHC`<m?BUau@N_O>Yct.MXo[>r5^uV&26@MkYB'Kiu\Y
K(*}ldO:ZQnI8t989fi+

输出应为:

test1
test2

我尝试使用下面的代码grep "[0-9][a-zA-Z]\+\.[a-zA-Z]\+[0-9]"查找字符串.现在我可以用cut command剪切出第一行，但是最后两行呢? 我认为使用AWK可以轻松解决我的问题，但我不知道如何解决. 谢谢

I have tried with following code grep "[0-9][a-zA-Z]\+\.[a-zA-Z]\+[0-9]" to find the string.Now i can cut the firs two line with cut command but last two line? I think with AWK can i solve my problem easily but i dont know how. Thanks

推荐答案

查看此行是否有帮助:

kent$  grep -Po '(?<=9S).*?(?=D9)' file
test1.test2

或更动态，这适用于您的示例:

or more dynamic, this works for your example:

kent$  grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' file
test1.test2

编辑

如@devnull所建议的那样，要获得所需的输出，您可以

as @devnull suggested, to get the desired output, you could

grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' file|tr '.' '\n'

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