使用java regex从字符串中提取版本

I could not seem to figure out the proper regex (java style) to extract just the version part for these string

我似乎无法弄清楚正确的正则表达式（java风格）只提取这些字符串的版本部分

EXPECTATION
spring-aop-4.2.5.RELEASE.jar  -> 4.2.5
rumi-1.js                     -> 1
BouncyCastle-Net-12-1.dll     -> 12-1

With the following java regex I keep getting a period at the end of line

使用以下java正则表达式，我会在行尾添加句点

\\b\\d[\\d|\\.|\\-]*\\b

Anyone can suggest a better regex?

任何人都可以建议更好的正则表达式？

 FAULTY_RESULT
 spring-aop-4.2.5.RELEASE.jar  -> 4.2.5.
 rumi-1.js                     -> 1.
 BouncyCastle-Net-12-1.dll     -> 12-1.

3 个解决方案

#1

Digit(s), followed by zero or more lots of a dot/dash and digit(s), not preceded by a word character:

数字，后跟零个或多个点/破折号和数字，前面没有单词字符：

(?<!\w)\d+([.-]\d+)*

In java, it can be done in one line:

在java中，它可以在一行中完成：

String version = packageName.replaceAll(".*?((?<!\\w)\\d+([.-]\\d+)*).*", "$1");

Here, the target term is captured while the regex matches the entire input and the replacement term returns the target (via the captured group).

这里，当正则表达式匹配整个输入并且替换项返回目标（通过捕获的组）时捕获目标项。

#2

Probably you can get this expression better, but this is my first attempt. Note. you need to scape this expression.

也许你可以更好地表达这种表达，但这是我的第一次尝试。注意。你需要改变这种表达方式。

(\d+(\.?|-?|\d)+\d|\d)

regards.

问候。

#3

-1

Try something like this:

尝试这样的事情：

\b\d[\d|.|\-]*\d\b

The doubling of the back-slashes goes into the String constant only:

反斜杠的加倍仅进入String常量：

"\\b\\d[\\d|.|\\-]+*\\d\\b"

However this will not match a single digit version...

但是，这与单个数字版本不匹配......

#1