regex - Regular expression which matches at least two words from a list

Question

I have a list of words - "foo", "bar", "baz" - and I want to write a regexp which would match strings which contain at least 2 of them. E.g., "foo baz" should match while "ba foo z" should not.

The obvious solution "(foo|bar|baz).*(foo|bar|baz)" works, but I find it unsatisfactory because it lists the words twice. What if I have 25 words instead of just 3? What if I am looking for strings which contain at least 4 given words instead of just 2?

Answer 1

听起来您不是在寻找确切的单词，因此 Donkey 的解决方案可能不是您想要的

((foo|bar|baz).*?){2}

它在文本中搜索任何这些字符串，然后搜索任何字符，直到再次找到这些可选字符串之一，并且由于惰性任何字符部分将通过不匹配任何内容来完成，因此匹配完成。

如果您希望它匹配多行，请务必打开 dot all，或使用 \s\S 而不是 dot。

Answer 2

我认为这个解决方案应该有效：

"(foo|bar|baz).*\s+\1(\s+|$)"

这\s意味着需要一个空格字符来确保您找到确切的单词而不仅仅是前缀。例如，"foo ... fooo"无法识别。