How to use regex lookahead and match the previous string / character class

问题

Trying to use negative look ahead to match a number if it doesn't precede a % sign.

\d+(?!%) .
8989% .
//matches 898, but not the 9% .
I'd like it to match 8989 as a whole.

Also is it possible use negative look ahead with matching a whole character class or more complex regex?
[\d.+](?!%) .
\d+(\.\d{1,2})?(?!%) .
Which would match decimals not preceding a %

回答1:

The \d+(?!%) pattern matches one or more digits, and grabs 8989 in 8989% at first, but the (?!%) negative lookahead fails that match, and the engine, seeing the + quantifier, starts backtracking. It discards the last 9 from the match buffer and retries the (?!%) lookahead that succeeds as 898 is not followed with % symbol.

You may use

/\d+(?![\d%])/g

See the regex demo

The (?![\d%]) negative lookahead will fail the match if 1+ digits is followed with any digit or % char, and thus will not return partial matches of 1+ digits that are followed with a % symbol.

来源：https://stackoverflow.com/questions/52469613/how-to-use-regex-lookahead-and-match-the-previous-string-character-class