I have an array of strings like so:
["Author Name, (p. 123). (2019). Company.", "Author Name, (p. 321). (2021). Company."]
How can I return the page numbers that start with and contain the pattern (p.?
So far I have tried /\(([^)\)]+)\)/ however it returns everything with parentheses. I only want the page numbers with parentheses.
You can match the p. before the capture group and capture the numbers. You don't have to escape the parenthesis in the character class, so you can remove \) and leave just )
\(p\.\s*([^)]+)\)
See a regex demo.
const regex = /\(p\.\s*([^)]+)\)/g;
[
"Author Name, (p. 123). (2019). Company.",
"Author Name, (p. 321). (2021). Company."
].forEach(s => {
Array.from(s.matchAll(regex), m => console.log(m[1]))
});I'd capture digits (you need (p.s.)?), case-insensitive (what about (P.12)?):
/\(p\.\s*(\d+)\)/ig
Code
const regex = /\(p\.\s*(\d+)\)/gi
const string = "Author Name, (p. 123). (2019). Company."
console.log(string.match(regex).map(x => x.replace(/\D+/g, '')))EXPLANATION
NODE EXPLANATION
--------------------------------------------------------------------------------
\( '('
--------------------------------------------------------------------------------
p 'p'
--------------------------------------------------------------------------------
\. '.'
--------------------------------------------------------------------------------
\s* whitespace (\n, \r, \t, \f, and " ") (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
\) ')'
See regex proof.