CODEWITHSUNDEEP
javaregex
Petr B
Petr B

Reputation: 529

Regex to check even number of preceding characters

I needed to ensure the string doesn't end with \' so used negative look behind:

\\:'(.+)(?<!\\\\)'

However, it could end with \\'. Basically, it could end with ' or ' preceded by the even number of backslashes \.

It is implemented in Java.

Upvotes: 0

Views: 354

Answers (2)

user557597
user557597

Reputation:

The only way

(?<!\\)(?:\\\\)*'$

Explained 

 (?<! \\ )          # Not an escape behind us, forces only even escapes ahead
 (?: \\ \\ )*       # Any amount of even escapes
 '                  # Quote 
 $                  # EOS

Notes - It is a fact that even escapes don't escape anything, so to enforce that only even escapes can be ahead, a negative look behind (?<!\\) is used.

Upvotes: 1

Andreas
Andreas

Reputation: 159086

Write it as a regex to find what you don't want, then include that regex in a zero-width negative lookahead. Description below uses xxx to indicate regex built up to that point.

You want to find \' at the end: \\'$

You want to include even number of \ before that. (?:\\\\)*xxx

You don't want a \ before that: (?<!\\)xxx

Anything before that is allowed: .*xxx

Embed in a negative lookahead then match anything: ^(?!xxx).*$

Make sure . matches line breaks (optional): (?s)xxx

All combined: (?s)^(?!.*(?<!\\)(?:\\\\)*\\'$).*$

As a Java literal: "(?s)^(?!.*(?<!\\\\)(?:\\\\\\\\)*\\\\'$).*$"

Demo on regex101.com shows that lines ending with odd number of \ before final ' is not selected, i.e. is invalid.

Upvotes: 0

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