Extracting lines consisting of three digits with Windows findStr

Question

when I am using the code in my windows CMD it is not giving me any output.

I am using findstr -E ^[0-9]{3}$ and it does not provide the expected output.

This is a sample file:

Your code specified the start and end of a line. So the regex will try to match a line containing only 3 digits. Maybe remove the `^` and `$` — Robo Mop, Feb 25 '20 at 09:22
[`findstr`](https://learn.microsoft.com/en-us/windows-server/administration/windows-commands/findstr) is not `grep`. — Wiktor Stribiżew, Feb 25 '20 at 09:35
`findstr /R "^[0-9][0-9][0-9]$"` (see: `findstr /?` for more info, and the link from @WiktorStribiżew ) — Luuk, Feb 25 '20 at 10:46

Wiktor Stribiżew · Answer 1 · 2020-02-25T10:50:51.383

Windows findStr is not grep, you need to use

findstr /r /c:"^[0-9][0-9][0-9]$" test.txt

The /r option makes findStr search for a "regex" pattern and /c:"..." uses the specified text as a literal search string.

Since the regex features findStr supports are very limited, you can't use {3}, you need to repeat the [0-9] thrice. Test:

C:\>findstr /r /c:"^[0-9][0-9][0-9]$" test.txt
834
519
645

Note: to make it extract 645, I had to make sure there is a trailing linebreak after this last line in the test.txt file.

1 Answers1