正则表达式从字符串的开头和结尾删除某些字符

Question

I'm trying to create a regex to remove all characters that are not space, tabs or alpha numeric from a string. Though if the character % exists and the start or end of the string they should be not be matched.

So far i have this

(?!<=^\%^)*[^\w\s](?<!%)

But it does not work as intented: https://regexr.com/5mckp

for example

%foo: bar [10]%
foo% bar%
foo%bar

should become

%foo bar 10%
foo bar%
foobar

Answer 1

Probably something like this should work:

(?!^\%*)[^\w\s](?<!\%$)

Your error stands in the fact that the sequence ^\%^ will never occur since you are asking that the start of the sequence is after a % sign. Also the universal quantifier belongs to the % sign and not to the negative group. You should also check for the end of the line $ . The rest was fine

Answer 2

[^\w\s] does not match an underscore which is not alphanumeric, you need [^\w\s]|_ .

Capture % at the start and end to keep them, use

string.replace(/(^%|%$)|[^\w\s]|_/g, '$1')

See proof . No lookbehind ensures the pattern work in Safari and other browsers not supporting lookbehind yet.

JavaScript code :

 const strings = ['%foo: bar___ [10]%','foo% bar%','foo%bar']; strings.forEach( string => console.log(string.replace(/(^%|%$)|[^\w\s]|_/g, '$1')) )