Java正则表达式负向前瞻不能正常工作

Question

I'm having a struggle creating a regex that verifies if a string does not end in a sequence of X characters (the same one).

For example: x=5

1xxxxx - should fail
1xxx - should pass

My regex so far : 1(?!\\\\w\\1{4}) does not work. Returns false in both cases.

Thanks

Answer 1

您需要添加字符串锚点结束：

1(?!.*(.)\1{4,}$).*$

Answer 2

This was a tough one, but if the condition in your question is correct, "string does not end in a sequence of X characters", this should do it:

(?:(.)(?:(?:(?!\1).){1,4})|^(.)\2{0,3})$

It's basically two possible alterations inside a non-capturing group (?: - ) .

The last part - ^(.)\\2{0,3} - simply tests if its a complete string consisting of only one character, less than five long.

The first part - (.)(?:(?:(?!\\1).){1,4}) - checks for a change of character by capturing one, and by a negative look-ahead making sure it doesn't match the next one, which is matched by the . and tested for a repetition of less than four times

The both alterations is tested for end of string $ outside the group.

Check it out here at regex101 .