正则表达式\\ b匹配具有特殊字符的字符串

Question

Having issues with \\b expression that shouldn't match with words with special characters.

I have this regex

\bstr=flavor\b

I was reading about escaping special characters but I'm not sure where to put it.

My goal is for the regex not to match the following string

str=flavor%20sweet --> Shouldn't match (1) 
str=flavorsweet --> shouldn't match (2)
declare str=flavor --> should match (3)

2 and 3 already works with the current regex but 1 I'm having issues with since it still matches where it should take into account the %

How should I change my regex to have it treat % as a character?

Answer 1

If your regex engine accept negative lookahead, you can use:

\bstr=flavor\b(?!%)

to force that the character directly following your str=flavor is not a % char.

demo: https://regex101.com/r/kSmLzW/1

Also you could add [ \\t]* around the = char to accept str = flavor for example.

This gives the regex: ( \\bstr[ \\t]*=[ \\t]*flavor\\b(?!%) )

For older versions of JS you can use:

\b(str=flavor)\b(?:[^%]|$)

demo: https://regex101.com/r/kSmLzW/2 where you can refer back to match via back reference if necessary. Also if there are other characters that you do not want to be taken into account in top of the % you can just add them in the character range exception definition [^]