How to make a regular expression

Question

Good evening everyone!
You need to write a regular expression that selects a line from the file and the lines following it, before the next key

For example, in the file a lot of different keys are described:

/P Вывод запросов перед созданием каждого нового файла. /S Копирование только непустых каталогов с подкаталогами. /E Копирование каталогов с подкаталогами, включая пустые. Эквивалентен сочетанию ключей /S /E. Совместим с ключом/T. /C Продолжение копирования вне зависимости от наличия ошибок.

I use the command FINDSTR /C: "/E " file.txt > lab.txt

With such an entry in the file will display the following line:

 /E Копирование каталогов с подкаталогами, включая пустые.

and I need to get it out:

 /E Копирование каталогов с подкаталогами, включая пустые. Эквивалентен сочетанию ключей /S /E. Совместим с ключом/T.

Ie to get 2 lines and if there are 3 of them in the key, then 3 is described

Question: how to change the regular expression? command key to change to FINDSTR /R "регулярное выражение" file.txt > file2.txt

Community spirit ♦ one · Accepted Answer · 2016-12-17T21:38:36

FINDSTR not able to search for multi-line matches (unless you refer to dark hacks ). It would be possible to use JREPL.bat , but, in my opinion, he is not on friendly terms with Russian. So I can advise pcregrep (there is a port under Windows pcregrep ):

 pcre2grep -M "^/E .+(\r?\n .+)*" file.txt

I downloaded grep.exe, threw it into System32, I launch the console, I go to the directory where my file is located, I write grep -M "^ / E. + (\ r? \ n. +) *" file.txt> text.txt, but it doesn't send anything to the file, the file is empty
Thank you very much, thanks to you, it turned out, I did this like grep -A 1 "/ E" file.txt> text.txt, where -A 1 is the number of lines after the found line to be output

How to make a regular expression

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