正则表达式将所有字符都放在第一个空格的右边? [英] Regex to get all character to the right of first space?

问题描述

我正在尝试制作一个正则表达式，该表达式将匹配(但不包括)字符串中第一个空格之后的所有字符.

I am trying to craft a regular expression that will match all characters after (but not including) the first space in a string.

输入文本:

foo bar bacon

想要的匹配:

bar bacon

到目前为止我发现的最接近的是:

The closest thing I've found so far is:

\s(.*)

然而，这匹配除了bar bacon"之外的第一个空格，这是不可取的.任何帮助表示赞赏.

However, this matches the first space in addition to "bar bacon", which is undesirable. Any help is appreciated.

推荐答案

您可以使用 positive lookbehind:

(?<=\s).*

(演示)

虽然看起来您已经在当前正则表达式中的 .* 周围放置了一个捕获组，因此您可以尝试抓取它.

Although it looks like you've already put a capturing group around .* in your current regex, so you could just try grabbing that.

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