正则表达式从字符串中提取版本

I have a string which has below value

我有一个低于值的字符串

String1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64

字符串1 = winos-APP-1.2.0-4.20120308.InLinuxOS.x86_64

I need to extract only version from this string which is "1.2.0-4"

我只需从这个字符串中提取“1.2.0-4”的版本

I have tried regular expression as mentioned below with sed

我用sed尝试了正则表达式，如下所述

sed -ne 's/[^0-9]*\(\([0-9]\.\)\{0,1\}[0-9][^.]\).*/\1/p'

but I am only getting result "1.2.0-", missing number after "-" which is 4. I tried correcting it but it is returning null.

但我只得到结果“1.2.0-”，在“ - ”之后缺少数字，这是4.我尝试纠正它但它返回null。

Kindly, advise

好心提醒

5 个解决方案

#1

how about grep:

怎么样grep：

grep -Po "(?<=-)[\d.-]*(?=.\d{8})"

#2

This should work, and should also account for any of the version numbers being more than one digit long.

这应该有效，并且还应该考虑长度超过一位数的任何版本号。

String1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64
sed -n 's/[^0-9]*\(\([0-9]\+\.\)\{0,2\}[0-9]\+-[0-9]\+\).*/\1/p' <<< "$String1"
1.2.0-4

BTW, this will also match version strings like (which from your question is not clear if you want this behavior or not):

顺便说一下，这也会匹配版本字符串（如果你想要这个行为，你的问题就不清楚了）：

1-1
1.2-1

If you want to enforce w.x.y-z, you could use this:

如果要强制执行w.x.y-z，可以使用：

sed -n 's/[^0-9]*\(\([0-9]\+\.\)\{2\}[0-9]\+-[0-9]\+\).*/\1/p' <<< "$String1"

#3

sed -n 's/^.*-\([0-9.]*-[0-9]*\)\..*$/\1/p'

#4

Can you try this?

你能试试吗？

[^0-9]*\(\([0-9]\.\)\{0,2\}[0-9][^.]\).*

[^ 0-9] * \（\（[0-9] \。\）\ {0,2 \} [0-9] [^。] \）。*

#5

You could also try parameter expansion:

您还可以尝试参数扩展：

string1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64
string2=${string1%".${string1#*-*-*-*.}"}
version=${string2#*-*-}
printf "%s\n" "$version"

#1