JS 正则表达式匹配以字符开头的单词，但如果它是双精度则忽略

Question

So I have this regular exp :: /(?:^|\W)\£(\w+)(?!\w)/g

This is meant to match words following the character £ . ie; if you type Something £here it will match £here .

However, if you type ££here I don't want it to be matched, however, i'm unsure how to match £ starting but ignore if it's ££ .

Is there guidance on how to achieve this?

Answer 1

You can add £ to \W :

/(?:^|[^\w£])£(\w+)/g

Actually, (?!\w) is redundant here (as after a word char, there is no more word chars) and you can remove it safely.

See the regex demo . Details :

• (?:^|[^\w£]) - start of string or any single char other than a word and a £ char
• £ - a literal char
• (\w+) - Group 1: one or more word chars.

Answer 2

If you also don't want to match $£here

(?<!\S)£(\w+)

Explanation

• (?<!\S) Assert a whitespace boundary to the left
• £ Match literally
• (\w+) Capture 1+ word characters in group 1

See a regex101 demo .

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If a lookbehind is not supported:

(?:^|\s)£(\w+)

See another regex101 demo .