Grep for line having only 2 or 3 digits

Question

I'm trying to print line containing 2 or 3 numbers along with the rest of the line. I came with the code:

grep -P '[[:digit:]]{2,3}' address

But this even prints the line having 4 digits. I don know why is this happening.

Output: Neither this code works;

grep -E '[0-9]{2,3}' address

Here is the file containing address text:

12 main st

123 main street

1234 main street

I have already specified to print 2 or 3 values with {2,3} still the filter doesn't work and more than 3 digits line is being printed. Can anyone assist me on this? Thank you so much.

Try this: `^[[:digit:]]{2,3}\b` (this works if line start with number) — Alireza, Sep 22 '21 at 05:49
It's word boundary, see [this](https://stackoverflow.com/questions/1324676/what-is-a-word-boundary-in-regex-does-b-match-hyphen) for more about the word boundary. — Alireza, Sep 22 '21 at 05:56
Another solution if after numbers always is a space: `^[[:digit:]]{2,3} ` — Alireza, Sep 22 '21 at 05:58
By way of explanation, lines that contain 4-digit sequences, by definition, necessarily contain 3-digit sequences, so `grep` will find them too. You need to search for *"not a digit followed by 2 or 3 digits followed by not a digit"*. — Mark Setchell, Sep 22 '21 at 07:26

score 2 · Answer 1 · answered Jan 03 '22 at 11:02

You can use inverted grep (-v) to filter all lines with 4 digits (and above):

grep -vE '[0-9]{4}' address

EDIT:

I noticed you want only 2 or 3 digit along the line, so first command will get you also 1 digit.

Here's the fix, again using same method:

grep -E '[0-9]{2,3}' txt.txt | grep -vE '[0-9]{4}'

Grep for line having only 2 or 3 digits

1 Answers1