匹配非空白字符的正则表达式 [英] Regular Expression to Match Non-Whitespace Characters

问题描述

我需要制作一个匹配以下内容的正则表达式:

I need to make a regular expression that matches something like:

JG2144-141/hello

或

!

但不是:

laptop bag

或仅由空白字符组成的字符串 (' ').

or a string consisting of whitespace chars only (' ').

现在我有 [A-Za-z0-9-!/\S]，但它不起作用，因为它仍然单独与笔记本电脑和包匹配.它根本不应该匹配笔记本电脑包和空字符串.

Right now I have [A-Za-z0-9-!/\S], but it isn't working because it still matches with laptop and bag individually. It shouldn't match laptop bag and the empty string at all.

推荐答案

[A-Za-z0-9-!/\S]中的\S使得此字符类等于 \S，但您要确保字符串中的所有字符都是非空白字符.这就是为什么你应该用 ^ 和 $ 锚包裹模式并在 \S 之后添加一个 + 量词到匹配 1 个或多个此子模式.

The \S in [A-Za-z0-9-!/\S] makes this character class equal to \S, but you want to make sure all chars in the string are non-whitespace chars. That is why you should wrap the pattern with ^ and $ anchors and add a + quantifier after \S to match 1 or more occurrences of this subpattern.

你可以使用

^\S+$

查看正则表达式演示

详情

• ^ - 字符串的开始
• \S+ - 1 个或多个非空白字符
• $ - 字符串结束.

• ^ - start of string
• \S+ - 1 or more non-whitespace chars
• $ - end of string.

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