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.
CodePudding user response:
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);
CodePudding user response:
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(. )/
CodePudding user response:
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
CodePudding user response:
You don't have to use replace. You can optionally match a digit and a space at the start of the string with ^(?:\d )?
and capture the rest of the line in group 1 that starts with a letter.
Note that if you want to use \s
that is could also match a newline.
^(?:\d )?([A-Za-z].*)
const regex = /^(?:\d )?([A-Za-z].*)/;
["1 Foo", "2 Foo bar", "Test", "3 A", "-test-"]
.forEach(s => {
const m = s.match(regex);
if (m) console.log(m[1])
});