Regex to match the letter string group between 2 numbers - regex

Question

Is it possible to match only the letter from the following string?
RO41 RNCB 0089 0957 6044 0001 FPS21098343
What I want: FPS
What I'm trying LINK : [0-9]{4}\s*\S+\s+(\S+)
What I get: FPS21098343
Any help is much appreciated! Thanks.

Answer 1

You can try with this:
var String = "0258 6044 0001 FPS21098343";
var Reg = /^(?:\d{4} )+ *([a-zA-Z]+)(?:\d+)$/;
var Match = Reg.exec(String);
console.log(Match);
console.log(Match[1]);

Answer 2

You can match up to the first one or more letters in the following way:
^[^a-zA-Z]*([A-Za-z]+)
^.*?([A-Za-z]+)
^[\w\W]*?([A-Za-z]+)
(?s)^.*?([A-Za-z]+)
If the tool treats ^ as the start of a line, replace it with \A that always matches the start of string.
The point is to match
^ / \A - start of string
[^a-zA-Z]* - zero or more chars other than letters
([A-Za-z]+) - capture one or more letters into Group 1.
The .*? part matches any text (as short as possible) before the subsequent pattern(s). (?s) makes . match line break chars.
Replace A-Za-z in all the patterns with \p{L} to match any Unicode letters. Also, note that [^\p{L}] = \P{L}.

Answer 3

To grep all the groups of letters that go in a row in any place in the string you can simply use:
([a-zA-Z]+)

Answer 4

You could use a capture group to get FPS:
\b[0-9]{4}\s+\S+\s+([A-Z]+)
The pattern matches:
\b[0-9]{4} A wordboundary to prevent a partial match, and match 4 digits
\s+\S+\s+ Match 1+ non whitespace chars between whitespace chars
([A-Z]+) Capture group 1, match 1+ chars A-Z
Regex demo
If the chars have to be followed by digits till the end of the string, you can add \d+$ to the pattern:
\b[0-9]{4}\s+\S+\s+([A-Z]+)\d+$
Regex demo