Question

I need to check if a string contains anything except the digit. But I’ve got a problem.

echo bdfbdfd | findstr /R [^0123456789]

returns nothing, but it should return bdfbdfd.
But

echo 123 | findstr /R [^0123456789]

returns ‘123’. Why? It should work vice versa.

Answer 1

After a lot of tests here we go:

echo 123 | findstr /R "[^0123456789]"
        ^ this space is what is matched, remove it and it won't match anything

Another way to do this is (as Christian.K stated) escaping the ^ thus writing

echo 123| findstr /R [^^0123456789]

(note the ^^)

For the bdfbdfd part putting the second regex between "" and removing the space makes it work for the specified string.

EDIT:

Note that you are matching a single character. In the match

echo bdfbdfd | findstr /R [^0123456789]

nothing was matched because your string didn’t contain any digit (not counting the ^ which needed to be escaped). In

echo 123 | findstr /R [^0123456789]

you were matching the first digit. With your regex you’ll match any string that contains a single match, so also trying to match 123(space) with the regex [^^0123456789] will output the whole string as a match because of the non-digit character in the end of the string.