What is the correct way to format this grep?

Question

I am having trouble figuring out where to place the \ in this command.

grep "^D.*\(A1|A2|A3\)$" input.txt > output.txt

I'm searching for each line that starts with a D and ends with A1, A2, or A3, which is at the end of the line.

For this particular case, you could use a simple character range `^D.*A[1-3]$` — steeldriver, Jan 26 '19 at 02:09

score 5 · Answer 1 · answered Jan 25 '19 at 23:33

5

You need to escape the pipes (|).

$ grep "^D.*\(A1\|A2\|A3\)$" <(printf 'D%s\n' A1 A2 A3)
DA1
DA2
DA3

Or use option -E for extended regex, then you don't need to escape anything.

$ grep -E "^D.*(A1|A2|A3)$" <(printf 'D%s\n' A1 A2 A3)
DA1
DA2
DA3

answered Jan 25 '19 at 23:33

wjandrea

We can use also `-P` instead of `-E` and this is really useful in some cases, see: [Grep -E, Sed -E - low performance, but why?](https://askubuntu.com/q/960557/566421) – pa4080 Jan 26 '19 at 14:36

score 3 · Accepted Answer · answered Jan 25 '19 at 23:36

3

You just needed to escape the or bars - you were almost there:

grep "^D.*\(A1\|A2\|A3\)$"

Note you can also use egrep instead of all the escapes:

egrep "^D.*(A1|A2|A3)$"

answered Jan 25 '19 at 23:36

Eric Mintz

pa4080 · Answer 3 · 2019-01-26T14:26:01.017

0

Here is sed implementation, that uses the combination of the option -n and the command p:

sed -rn '/^D.*(A1|A2|A3)$/p' in-file

sed -n '/^D.*\(A1\|A2\|A3\)$/p' in-file

sed -n '/^D.*A[1-3]$/p' in-file

option -r, --regexp-extended: use extended regular expressions in the script.
option -n, --quiet, --silent: suppress automatic printing of pattern space.
command p: print the current pattern space.

edited Jan 26 '19 at 14:26

answered Jan 26 '19 at 13:37

pa4080

3 Answers3