如何在Java中使用条件编写正则表达式?

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英文:

How to compose regex with condition in Java?

问题

条件

  • 数字可以是整数或小数(共8位数字)
  • 数字只能由空格(一个或多个)和逗号(0个或一个)分隔
  • 逗号只能在数字内部(一行开头和结尾不能有逗号)
  • 一行开头可能有空格(一个或多个)
  • 一行结尾可能有空格(一个或多个)

我的做法

([\s]*\d+(\.{1}\d+)?[\s\,\s]+){7}(\d+(\.{1}\d+)?[\s]*){1}
                    ^ 这里   ^

这基本上是正确的,但有一个例外情况。在以下字符串中,我得到的是true,但需要的是false

String s1 = " 0 , 4.4 3.2,, 4.1      2 4 1 7.7";

我不能这样做:

> 数字之间可以用一个逗号分隔,或者没有逗号,但在这种情况下,必须有一个空格(一个或多个)。

英文:

Condition:

  • numbers can be either int or double (total 8 digits)
  • numbers can only be separated by spaces (one or many) and commas (0 or one)
  • commas can only be inside numbers (there can be no commas at the beginning and end of a line)
  • there may be spaces at the beginning of the line (one or many)
  • there may be spaces at the end of the line (one or many)

What I do:

([\s]*\d+(\.{1}\d+)?[\s\,\s]+){7}(\d+(\.{1}\d+)?[\s]*){1}
                    ^ this   ^

That's ok, except for one condition.
On this string I get true, but need false:

String s1 = " 0 , 4.4 3.2,, 4.1      2 4 1 7.7";

I can't do this:

> Numbers can be separated by only one comma or no comma, but in this case there must be a space (one or many).

答案1

得分: 1

这是我针对空格/逗号情况提出的解决方案(还删除了一些多余部分):

Pattern.compile("([\\s]*\\d+(\\.\\d+)?(?:\\s*,\\s*|\\s+)){7}(\\d+(\\.\\d+)?[\\s]*)");

根据你提供的示例,似乎可以达到你的要求。(你在使用 {1} 的地方通常是隐含的,所以我去掉了这些)

英文:

This is the one I came up with for the spaces/commas (also removed some redundant parts):

Pattern.compile("([\\s]*\\d+(\\.\\d+)?(?:\\s*,\\s*|\\s+)){7}(\\d+(\\.\\d+)?[\\s]*)");

Seems to do what you want with the sample you provided, at least. (The use of {1} is generally implied for the places you used it, so I removed those)

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  • 本文由 发表于 2020年8月30日 06:58:08
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