使用 bash 打印具有相同模式的两行之间的行

Question

I have a text file containing with string pattern 'OG' -

OG
ANNNFKSODJFHFJ
SSJSJKSKJSAJAS
SSSSSSSSSSSSFA
OG
FALJFNAFAFNAFJL
AFJLJSLJFLFSLFL
ASJFAJFAKFKAFKK
OG
AJSFLJASFLSFLFF
SJFLAFLAFLFLAFA
ASASFASFOFLJAJF

I want to print the lines between the first two lines that has 'OG' as string pattern, ie The result should be -

ANNNFKSODJFHFJ
SSJSJKSKJSAJAS
SSSSSSSSSSSSFA

Here, any suggestions using 'sed' or 'awk'. I don't want to use printing by line number but rather with pattern search.

I tried using awk, but not working -

awk '/^OG/{flag=1;next}/^OG/{flag=0}flag' file.txt

Answer 1

You're almost there:

$ awk '/^OG/ {if(go) exit; go=1; next} go {print}' file.txt
ANNNFKSODJFHFJ
SSJSJKSKJSAJAS
SSSSSSSSSSSSFA