多次匹配任何字符，但在给定字符处停止 [英] Match any characters more than once, but stop at a given character

问题描述

我正在编写一个正则表达式，用于识别字符串中的命令.我有三个可能的命令开头的单词，它们总是以分号结尾.

I am writing a regex that will be used for recognizing commands in a string. I have three possible words the commands could start with and they always end with a semi-colon.

我相信正则表达式应该是这样的:

I believe the regex pattern should look something like this:

(command1|command2|command3).+;

我发现的问题是，由于 . 匹配任何字符并且 + 告诉它匹配一个或多个，它会直接跳过 a 的第一个实例分号并继续.

The problem, I have found, is that since . matches any character and + tells it to match one or more, it skips right over the first instance of a semi-colon and continues going.

有没有办法让它在遇到分号的第一个实例处停止?除了 . 之外，还有什么我应该使用的东西吗?

Is there a way to get it to stop at the first instance of a semi-colon it comes across? Is there something other than . that I should be using instead?

推荐答案

您面临的问题是:(command1|command2|command3).+; 是 + 是贪婪的，这意味着它将匹配所有内容，直到最后一个值.

The issue you are facing with this: (command1|command2|command3).+; is that the + is greedy, meaning that it will match everything till the last value.

要解决此问题，您需要使其不贪婪，为此您需要添加 ? 运算符，如下所示:(command1|command2|command3).+?;

To fix this, you will need to make it non-greedy, and to do that you need to add the ? operator, like so: (command1|command2|command3).+?;

正如仅供参考，这同样适用于 * 运算符.添加一个 ? 将使它不贪婪.

Just as an FYI, the same applies for the * operator. Adding a ? will make it non greedy.

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