Linux 在shell中，用点作为分隔符分割字符串的一部分
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In shell, split a portion of a string with dot as delimiter
提问by rodee
I am new to shell scripting, can you please help with below requirement, thanks.
我是 shell 脚本的新手，请您帮忙解决以下要求，谢谢。
$AU_NAME=AU_MSM3-3.7-00.01.02.03
#separate the string after last "-", with "." as delimiter
#that is, separate "00.01.02.03" and print/save as below.
major=00
minor=01
micro=02
build=03
采纳答案by Mark Reed
First, note that you don't use $when assigning to a parameter in the shell. Your first line should be just this:
首先，请注意$在 shell 中分配参数时不要使用。你的第一行应该是这样的：
AU_NAME=AU_MSM3-3.7-00.01.02.03
Once you have that, then you can do something like this:
一旦你有了它，那么你可以做这样的事情：
IFS=. read major minor micro build <<EOF
${AU_NAME##*-}
EOF
where the ##*-strips off everything from the beginning of the string through the last '-', leaving just "00.01.02.03", and the IFS (Internal Field Separator) variable tells the shell where to break the string into fields.
其中##*-从字符串的开头到最后一个 '-' 去掉所有内容，只留下“00.01.02.03”，并且 IFS（内部字段分隔符）变量告诉 shell 在何处将字符串分解为字段。
In bash, zsh, and ksh93+, you can get that onto one line by shortening the here-document to a here-string:
在 bash、zsh 和 ksh93+ 中，您可以通过将 here-document 缩短为 here-string 来将其放到一行中：
IFS=. read major minor micro build <<<"${AU_NAME##*-}"
More generally, in those same shells (or any other shell that has arrays), you can split into an arbitrarily-sized array instead of distinct variables. This works in the given shells:
更一般地，在那些相同的 shell（或任何其他具有数组的 shell）中，您可以拆分为任意大小的数组而不是不同的变量。这适用于给定的外壳：
IFS=. components=(${AU_NAME##*-})
In older versions of ksh you can do this:
在旧版本的 ksh 中，您可以这样做：
IFS=. set -A components ${AU_NAME##*-}
That gets you this equivalence (except in zsh, which by default numbers the elements 1-4 instead of 0-3):
这让你得到这个等价（除了在 zsh 中，默认情况下元素编号为 1-4 而不是 0-3）：
major=${components[0]}
minor=${components[1]}
micro=${components[2]}
build=${components[3]}
回答by paddy
In bash, you can do something like this:
在 中bash，您可以执行以下操作：
version=$(echo $AU_NAME | grep -o '[^-]*$')
major=$(echo $version | cut -d. -f1)
minor=$(echo $version | cut -d. -f2)
micro=$(echo $version | cut -d. -f3)
build=$(echo $version | cut -d. -f4)
The grepcall uses -owhich outputs only the matching part of the line. The match itself is every non-hyphen character to the end of the line.
该grep调用使用-owhich 只输出该行的匹配部分。匹配本身是行尾的每个非连字符字符。
The cutcommand uses the delimeter .(-d.), and uses -fto select individual fields.
该cut命令使用分隔符.( -d.)，并用于-f选择单个字段。
It's a little clunky. I'm sure there are probably better ways to achieve this, but you can do quite a lot with grepand cutalone so they're handy tools to have in your arsenal.
它有点笨重。我敢肯定，有可能是更好的方式来实现这一点，但你可以做了很多与grep和cut单独所以他们得心应手的工具到您的阿森纳。
回答by choroba
You can use parameter expansion and the special IFS variable.
您可以使用参数扩展和特殊的 IFS 变量。
#! /bin/bash
AU_NAME=AU_MSM3-3.7-00.01.02.03
IFS=. VER=(${AU_NAME##*-})
for i in {0..3} ; do
echo ${VER[i]}
done
major=${VER[0]}
minor=${VER[1]}
micro=${VER[2]}
build=${VER[3]}
BTW, in an assignment, do not start the variable on the left hand side with a dollar sign.
顺便说一句，在赋值中，不要以美元符号开始左侧的变量。
