Regex to get one character not followed with digit

Question

How can I get first character if not have int inside:

I need to look all the place have '[' without integer after.

For example:

[abc] pass
[cxvjk234] pass
[123] fail

Right now, I have this:

((([[])([^0-9])))

It gets the first 2 characters while I need only one.

Answer 1

In general, to match some pattern not followed with a digit, you need to add a (?!\d) / (?![0-9]) negative lookahead to the expression:

\[(?!\d)
\[(?![0-9])
  ^^^^^^^^^

See the regex demo. This matches any [ symbol that is not immediately followed with a digit.

---

Your current regex pattern is overloaded with capturing groups, and if we remove those redundant ones, it looks like (\[)([^0-9]) - it matches a [ and then a char other than an ASCII digit.

You may use

(?<=\[)\D

or (if you want to only match the ASCII digits with the pattern only)

(?<=\[)[^0-9]

See the regex demo

Details:

• (?<=\[) - a positive lookbehind requiring a [ (but not consuming the [ char, i.e. not returning it as part of the match value) before...
• \D / [^0-9] - a non-digit. NOTE: to only negate ASCII digits, you may use \D with the RegexOptions.ECMAScript flag.