Regex to match a whole string only if it lacks a given substring/suffix

问题

I've searched for questions like this, but all the cases I found were solved in a problem-specific manner, like using !g in vi to negate the regex matches, or matching other things, without a regex negation.

Thus, I'm interested in a “pure” solution to this:

Having a set of strings I need to filter them with a regular expression matcher so that it only leaves (matches) the strings lacking a given substring. For example, filtering out "Foo" in:

Boo
Foo
Bar
FooBar
BooFooBar
Baz

Would result in:

Boo
Bar
Baz

I tried constructing it with negative look aheads/behinds (?!regex)/(?<!regex), but couldn't figure it out. Is that even possible?

回答1:

Try this regular expression:

^(?:(?!Foo).)*$

This will consume one character at a time and test if there is no Foo ahead. The same can be done with a negative look-behind:

^(?:.(?<!Foo))*$

But you can also do the same without look-around assertions:

^(?:[^F]*|F(?:$|[^o].|o(?:$|[^o])))*$

This matches any character except F or an F that is either not followed by a o or if followed by an o not followed by another o.

来源：https://stackoverflow.com/questions/1968800/regex-to-match-a-whole-string-only-if-it-lacks-a-given-substring-suffix