bash bash中冒号后的正则表达式提取字符串
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regular expression extract string after a colon in bash
提问by prolink007
I need to extract the string after the :in an example below:
我需要:在下面的示例中提取字符串：
package:project.abc.def
package:project.abc.def
Where i would get project.abc.defas a result.
project.abc.def结果我会得到什么。
I am attempting this in bash and i believe i have a regular expression that will work :([^:]*)$.
我正在 bash 中尝试这个，我相信我有一个可以工作的正则表达式:([^:]*)$。
In my bash script i have package:project.abc.defas a variable called apk. Now how do i assign the same variable the substring found with the regular expression?
在我的 bash 脚本中，我有package:project.abc.def一个名为apk. 现在我如何分配与正则表达式找到的子字符串相同的变量？
Where the result from package:project.abc.defwould be in the apkvariable. And package:project.abc.defis initially in the apkvariable?
凡从结果package:project.abc.def将在apk变量。并且package:project.abc.def最初是在apk变量中？
Thanks!
谢谢！
回答by gvalkov
There is no need for a regex here, just a simple prefix substitution:
这里不需要正则表达式，只需简单的前缀替换：
$ apk="package:project.abc.def"
$ apk=${apk##package:}
project.abc.def
The ## syntax is one of bash's parameters expansions. Instead of #, % can be used to trim the end. See this sectionof the bash man page for the details.
## 语法是 bash 的参数扩展之一。可以使用 % 代替 # 来修剪结尾。有关详细信息，请参阅bash 手册页的这一部分。
Some alternatives:
一些替代方案：
$ apk=$(echo $apk | awk -F'package:' '{print }')
$ apk=$(echo $apk | sed 's/^package://')
$ apk=$(echo $apk | cut -d':' -f2)
回答by Alex Merelli
$ string="package:project.abc.def"
$ apk=$(echo $string | sed 's/.*\://')
".*:" matches everything before and including ':' and then its removed from the string.
".*:" 匹配包含 ':' 之前的所有内容，然后将其从字符串中删除。
回答by chepner
Capture groups from regular expressions can be found in the BASH_REMATCHarray.
可以在BASH_REMATCH数组中找到来自正则表达式的捕获组。
[[ $str =~ :([^:]*)$ ]]
# 0 is the substring that matches the entire regex
# n > 1: the nth parenthesized group
apk=${BASH_REMATCH[1]}
