Bash, grep between two lines with specified string

Yep, normal grep won't do this. But grep with -P parameter will do this job.

$ grep -ozP '(?s)test1\n\K.*?(?=\ntest2)' file
abc
cvb
bnm

\K discards the previously matched characters from printing at the final and the positive lookahead (?=\ntest2) asserts that the match must be followed by a \n newline character and then test2 string.

Print from test1 to test2 (Trigger lines included)

awk '/test1/{f=1} /test2/{f=0;print} f'
awk '/test1/{f=1} f; /test2/{f=0}' 
awk '/test1/,/test2/'

test1
abc
cvb
bnm
test2

---

Prints data between test1 to test2 (Trigger lines excluded)

awk '/test1/{f=1;next} /test2/{f=0} f' 
awk '/test2/{f=0} f; /test1/{f=1}' 

abc
cvb
bnm

You could use sed:

sed -n '/test1/,/test2/p' filename

In order to exclude the lines containing test1 and test2, say:

sed -n '/test1/,/test2/{/test1/b;/test2/b;p}' filename

If you can only use grep:

grep -A100000 test1 file.txt | grep -B100000 test2 > new.txt

grep -A and then a number gets the lines after the matching string, and grep -B gets the lines before the matching string. The number, 100000 in this case, has to be large enough to include all lines before and after.

If you don't want to include test1 and test2, then you can remove them afterwards by grep -v, which prints everything except the matching line(s):

egrep -v "test1|test2" new.txt > newer.txt

or everything in one line:

grep -A100000 test1 file.txt | grep -B100000 test2 | egrep -v "test1|test2" > new.txt