这个正则表达式

^\w+(\s\w+)*$

将只允许单词之间有一个空格，不允许开头或结尾空格。

正则表达式解释如下:

^ Assert position at start of the string \w+ Match any word character [a-zA-Z0-9_] Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy] 1st Capturing group (\s\w+)* Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy] \s Match any white space character [\r\n\t\f ] \w+ Match any word character [a-zA-Z0-9_] Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy] $ Assert position at end of the string

如果你正在使用JavaScript，那么你可以使用这个正则表达式:

/^[a-z0-9_.-\s]+$/i

例如:

/^[a-z0-9_.-\s]+$/i.test("") //false
/^[a-z0-9_.-\s]+$/i.test("helloworld") //true
/^[a-z0-9_.-\s]+$/i.test("hello world") //true
/^[a-z0-9_.-\s]+$/i.test("none alpha: ɹqɯ") //false

这个正则表达式的唯一缺点是字符串完全由空格组成。" "也会显示为true。

试一试:

^(\w+ ?)*$

解释:

\w             - alias for [a-zA-Z_0-9]
"whitespace"?  - allow whitespace after word, set is as optional

我假设您不需要前导/尾随空格。这意味着你必须把正则表达式分成“第一个字符”，“中间的东西”和“最后一个字符”:

^[a-zA-Z0-9_][a-zA-Z0-9_ ]*[a-zA-Z0-9_]$

或者如果你使用类似perl的语法:

^\w[\w ]*\w$

另外:如果你故意措辞你的正则表达式，它也允许空字符串，你必须使整个事情可选:

^(\w[\w ]*\w)?$

如果你想只允许单个空格字符，它看起来有点不同:

^((\w+ )*\w+)?$

这个匹配0..N个单词后面跟着一个空格，加上一个没有空格的单词。并使整个东西可选，允许空字符串。

只需在正则表达式模式的末尾添加一个空格，如下所示:

[a-zA-Z0-9_ ]

试试这个吧:

result = re.search(r"\w+( )\w+", text)