Grep to display a couple lines until a specific text

Question

Lets say I have the following content in a text file

...
abc
123
qwerty
line0
...
line3
line10
max
....

How to I grep for abc and capture all lines after the match and stop at line that contains specific text, in this example max ?

Possible duplicate of [How to select lines between two patterns?](https://stackoverflow.com/questions/38972736/how-to-select-lines-between-two-patterns) — Sundeep, Oct 14 '17 at 03:50

score 2 · Answer 1 · answered Oct 13 '17 at 20:20

You can use sed:

sed '/abc/,/max/!d'

! is negation
d means delete
/addr1/,/addr2/ is an "address range"
The whole expression means "Delete the line unless it's between a line matching /abc/ and a line matching /max/".

Similarly in Perl:

perl -ne 'print if /abc/ .. /max/'

score 2 · Accepted Answer · answered Oct 13 '17 at 20:21

2

Even simpler than sed and my beloved Perl is awk.

awk '/abc/,/max/' filename

answered Oct 13 '17 at 20:21

Andy Lester

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score 0 · Answer 3 · answered Oct 13 '17 at 21:02

Line range addressing is simple in sed/awk, but if you really need to use grep, this will work:

grep -Poz 'abc[^\0]*max' file

The -z flag makes grep treat the input as null-byte terminated lines, so assuming your input doesn't contain any \0 bytes, the above PCRE will select everything between abc and max.

Grep to display a couple lines until a specific text

3 Answers3