Java正则表达式匹配无回车符，后跟换行符 [英] Java regular expression matching no carriage return followed by a linefeed

问题描述

我已经尝试过(^ \ r)\ n ，但这是行不通的.

I've tried (^\r)\n but this doesn't work.

您如何做到的?

(感谢您在类似Java的代码中使用(^ \\ r)\\ n )

(I appreciate that in Java-like code you need to use (^\\r)\\n)

谢谢

推荐答案

取决于您的要求:

• [^ \ r] \ n -换行符，除换行符外，前面带有任何字符.这意味着在换行符之前必须有一个字符，并且两个符号将被匹配.
• (?<！\ r)\ n -不带回车符的换行符.这意味着只有换行符号会被匹配，并且 \ r 仅会被测试是否存在(因为(?<！\ r)是一个负向后看，即是零宽度断言，该断言不占用任何文本，但是如果其中的模式恰好在字符串中的当前位置之前不存在，则返回true.

• [^\r]\n - linefeed that is preceded with any character but a carriage return. This means there must be a character before the linefeed and two symbols will be matched.
• (?<!\r)\n - linefeed that is not preceded with a carriage return. This means there only newline symbol will get matched and \r will only be tested for presence (as (?<!\r) is a negative lookbehind, a zero-width assertion that does not consume any text, but returns true if the pattern inside it is absent right before the current position in the string).

有关演示，请检查以下两个链接:

For a demo, please check these two links:

• 在 regex101.com ， (?<！\ r)\ n 匹配，因为换行符在网站上是纯 \ n
• 在 regexstorm.net 上，

• At regex101.com, (?<!\r)\n is matching as linebreaks are pure \n at the Web site
• At regexstorm.net, the same pattern does not match anything as linebreaks are \r\n there.

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