仅在单词唯一时匹配

Question

I've spent the day trying to use regex to match a word only where the word is unique. It's not working and I can't figure out what I'm doing wrong.

I need something that will match only a unique instance of "ST", such as:

• "1 MARY ST WASHINGTON"

but fail with these:

• "1 ST MARY ST WASHINGTON"
• "1 ST MARY ST MT ST HELENS"
• "1 MARY RD WE ST WASHINGTON"
• "1 MARY RD WASHINGTON"

I thought this lookahead would work, but no such luck:

(\bST\b)(?!(\bST\b))

What am I missing?

Answer 1

You can use RegExp /\\bST\\b/g and .match() , check .length of matched string

 const matches = (str, re = /\\bST\\b/g, match = str.match(re)) => !!match && match.length === 1; const arr = [ "1 MARY ST WASHINGTON" , "1 ST MARY ST WASHINGTON" , "1 ST MARY ST MT ST HELENS" , "1 MARY RD WEST WASHINGTON" , "1 MARY RD WASHINGTON" ]; console.log(arr.map(s => matches(s))); 

Answer 2

You can use following pattern:

^(?!(.*\bST\b){2,}).*\bST\b

This does a negative lookahead right away in which it checks if there are 2 or more occurrences of \\bST\\b . If there aren't, it moves on in the pattern and checks if there is at least one.