The target string looks like a number followed by a space and then followed by one or more letters, e.g. 1 Foo or 2 Foo bar.
I can use [^\d\s].+, but it doesn't work for single letters, e.g. 3 A.
What can be done here?
The workaround I use currently is to use replacing instead of matching.
from \d\s(.+)
to $1
But as a purist I prefer to use replacing if and only if we don't mean "replace something with nothing". When we need to replace something with nothing, I would prefer to use matching.
Just remove the square brackets. The square brackets alone indicate "any of this set", so you are matching either \d or \s. When you also add a ^ inside you are not indicating the beginning of the string, but you are negating the set. So, summing up, your regular expression means:
Match a single character that may be everything except a digit and a white space, then match everything.
If you remove the square brackets you will match \d followed by \s, and the ^ symbol will mean "beginning of the string".
/^\d\s(.+)/
I prefer using a regex replacement here:
var input = ["1 Foo", "2 Foo Bar", "No Numbers Here"];
var output = input.map(x => x.replace(/^\d+ /, ""));
console.log(output);If I didn't misread your question, this might be what you want:
exclude capture the number and space at the beginning
(?!^\d+)(?!\s+).*
This matches 1 Foo Bar to Foo Bar and 3 A to A