How to make sure that part of the pattern (keyword in this case) is in the pattern you're looking for, but it can appear in different places. I want to have a match only when it occurs at least once.

Regex:

 \b(([0-9])(xyz)?([-]([0-9])(xyz)?)?)\b

we only want the value if there is a keyword: xyz

Examples:

1. 1xyz-2xyz - it's OK
2. 1-2xyz - it's OK
3. 1xyz - it's OK
4. 1-2 - there should be no match, at least one xyz missing

I try positive lookup and lookbehind but this is not working in this case

CodePudding user response：

You can make use of a conditional construct:

\b([0-9])(xyz)?(?:-([0-9])(xyz)?)?\b(?(2)|(?(4)|(?!)))

See the regex demo. Details:

• \b - word boundary
• ([0-9]) - Group 1: a digit
• (xyz)? - Group 2: an optional xyz string
• (?:-([0-9])(xyz)?)? - an optional sequence of a -, a digit (Group 3), xyz optional char sequence
• \b - word boundary
• (?(2)|(?(4)|(?!))) - a conditional: if Group 2 (first (xyz)?) matched, it is fine, return the match, if not, check if Group 4 (second (xyz)?) matched, and return the match if yes, else, fail the match.

See the Python demo:

import re
text = "1. 1xyz-2xyz - it's OK\n2. 1-2xyz - it's OK\n3. 1xyz - it's OK\n4. 1-2 - there should be no match"
pattern = r"\b([0-9])(xyz)?(?:-([0-9])(xyz)?)?\b(?(2)|(?(4)|(?!)))"
print( [x.group() for x in re.finditer(pattern, text)] )

Output:

['1xyz-2xyz', '1-2xyz', '1xyz']

CodePudding user response：

Try this: \b(([0-9])?(xyz) ([-]([0-9]) (xyz) )?)\b Replace ? with Basically ?: zero or more and in your case you want to match one or more. Whih is