正则表达式:匹配 3 个连续的单词 [英] regex: matching 3 consecutive words

问题描述

我正在尝试查看一个字符串是否包含 3 个连续的单词(用空格分隔且没有数字)，但我构建的正则表达式似乎不起作用:

I'm trying to see if a string contains 3 consecutive words (divided by spaces and without numbers), but the regex I have constructed does not seem to work:

print re.match('([a-zA-Z]+\b){3}', "123 test bla foo")
None

这应该返回 true，因为字符串包含 3 个单词test bla foo".

This should return true since the string contains the 3 words "test bla foo".

实现这一目标的最佳方法是什么?

What is the best way to achieve this?

推荐答案

做:

(?:[A-Za-z]+ ){2}[A-Za-z]+

• (?:[A-Za-z]+ ){2}:未捕获组(?:[A-Za-z]+ ) 匹配一个或多个后跟空格的字母字符，{2} 匹配两个这样连续的组

  • (?:[A-Za-z]+ ){2}: the non-captured group (?:[A-Za-z]+ ) matches one or more alphabetic characters followed by space, {2} matches two such successive groups

    [A-Za-z]+ 匹配前两个单词后的一个或多个字母字符，成为第三个单词

    [A-Za-z]+ matches one or more alphabetic character after the preceding two words, making the third word

    演示

    如果您希望单词被任何空格分隔而不仅仅是空格:

    If you want the words to be separated by any whitespace instead of just space:

    (?:[A-Za-z]+\s){2}[A-Za-z]+
    

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