构建正则表达式以仅匹配 2 个单词 [英] Building regex to match 2 words only

问题描述

我正在尝试使正则表达式仅匹配 2 个单词和一个步调.没有特殊符号，只有[a-zA-Z]空格[a-zA-z].

I'm trying to make regexp that match only 2 words and a single pace between. No special symbols, only [a-zA-Z] space [a-zA-z].

Foo Bar      # Match    (two words and one space only)
Foo          # Mismatch (only one word)
Foo  Bar     # Mismatch (2 spaces)
Foo Bar Baz  # Mismatch (3 words)

推荐答案

你想要 ^[a-zA-Z]+\s[a-zA-Z]+$

^   # Matches the start of the string
+   # quantifier mean one or more of the previous character class 
\s  # matches whitespace characters
$   # Matches the end of the string

锚点 ^ 和 $ 在这里很重要.

The anchors ^ and $ are important here.

演示:

if "foo bar" =~ /^[a-zA-Z]+\s[a-zA-Z]+$/ 
    print "match 1"
end 
if "foo  bar" =~ /^[a-zA-Z]+\s[a-zA-Z]+$/ 
    print "match 2"
end 
if "foo bar biz" =~ /^[a-zA-Z]+\s[a-zA-Z]+$/ 
    print "match 3"
end 

输出:

Match 1

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