Java 正则表达式 - 匹配具有零个或一个空格的字符串

Regex - Match a string which has zero or one spaces

提问by crazyfool

I'm trying to match a string which starts with @, can contain any amount of letters or numbers but can only contain a maximum of one space (or zero spaces). So far I have

我试图匹配一个以@ 开头的字符串，可以包含任意数量的字母或数字，但最多只能包含一个空格（或零个空格）。到目前为止我有

@([A-Za-z0-9]+)

which matches the characters but without the space. I think I need \s{0,1} but I'm not sure where to put it.. Can anyone help?

匹配字符但没有空格。我想我需要 \s{0,1} 但我不知道把它放在哪里..有人可以帮忙吗？

Thanks.

谢谢。

采纳答案by Adam Yost

Assuming you only care about spaces in the word, not leading or trailing then you could use this:

假设你只关心单词中的空格，而不是前导或尾随，那么你可以使用这个：

@[A-Za-z0-9]* ?[A-Za-z0-9]*

Explanation:

解释：

@Starts with literal @

@以文字@开头

[A-Za-z0-9]Any letter or number

[A-Za-z0-9]任何字母或数字

*Letter or number can be length {0,infinity}

*字母或数字的长度可以是 {0,infinity}

?Space char, 0 or one times

?空格字符，0 或 1 次

[A-Za-z0-9]*Any number of trailing letters or spaces after the space (if there is one)

[A-Za-z0-9]*空格后任意数量的尾随字母或空格（如果有）

回答by Avinash Raj

You could try the below regex to match the words which starts with @follwed by any number of letters or numbers with an optional space,

您可以尝试使用下面的正则表达式来匹配@以任意数量的字母或数字开头的单词，并带有可选的空格，

^@[a-zA-Z0-9]+ ?[a-zA-Z0-9]*$

DEMO

演示

Java pattern would be,

Java模式将是，

"^@[a-zA-Z0-9]+ ?[a-zA-Z0-9]*$"

回答by anubhava

You can use this regex with negative lookahead:

您可以将此正则表达式与负前瞻一起使用：

^@((?!(?:\S* ){2})[A-Za-z0-9 ]+)$

回答by Braj

Is this what you needed?

这是你需要的吗？

@[A-Za-z0-9]+(?:\s?)[A-Za-z0-9]+

Online demo

在线演示