bash 如何在bash shell中将一个字符串拆分为多个变量？
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How to split one string into multiple variables in bash shell?
提问by crunchybutternut
I've been looking for a solution and found similar questions, only they were attempting to split sentences with spaces between them, and the answers do not work for my situation.
我一直在寻找解决方案并发现了类似的问题，只是他们试图用空格分隔句子，而答案不适用于我的情况。
Currently a variable is being set to something a string like this:ABCDE-123456
and I would like to split that into 2 variables, while eliminating the "-". i.e.:var1=ABCDEvar2=123456
目前，一个变量被设置为这样的字符串：ABCDE-123456
我想将其拆分为 2 个变量，同时消除“ -”。IE：var1=ABCDEvar2=123456
How is it possible to accomplish this?
怎么可能做到这一点？
This is the solution that worked for me:var1=$(echo $STR | cut -f1 -d-)var2=$(echo $STR | cut -f2 -d-)
这是对我有用的解决方案：var1=$(echo $STR | cut -f1 -d-)var2=$(echo $STR | cut -f2 -d-)
Is it possible to use the cutcommand that will work without a delimiter (each character gets set as a variable)?
是否可以使用无需分隔符的cut命令（每个字符都设置为变量）？
var1=$(echo $STR | cut -f1 -d?)var2=$(echo $STR | cut -f1 -d?)var3=$(echo $STR | cut -f1 -d?)etc.
var1=$(echo $STR | cut -f1 -d?)var2=$(echo $STR | cut -f1 -d?)var3=$(echo $STR | cut -f1 -d?)etc.
采纳答案by Rob I
Ifyour solution doesn't have to be general, i.e. only needs to work for strings like your example, you could do:
如果您的解决方案不必是通用的，即只需要为您的示例那样的字符串工作，您可以这样做：
var1=$(echo $STR | cut -f1 -d-)
var2=$(echo $STR | cut -f2 -d-)
I chose cuthere because you could simply extend the code for a few more variables...
我选择cut这里是因为您可以简单地为更多变量扩展代码......
回答by Paused until further notice.
readwith IFSare perfect for this:
read与IFS是完美的：
$ IFS=- read var1 var2 <<< ABCDE-123456
$ echo "$var1"
ABCDE
$ echo "$var2"
123456
Edit:
编辑：
Here is how you can read each individual character into array elements:
以下是如何将每个单独的字符读入数组元素：
$ read -a foo <<<"$(echo "ABCDE-123456" | sed 's/./& /g')"
Dump the array:
转储数组：
$ declare -p foo
declare -a foo='([0]="A" [1]="B" [2]="C" [3]="D" [4]="E" [5]="-" [6]="1" [7]="2" [8]="3" [9]="4" [10]="5" [11]="6")'
If there are spaces in the string:
如果字符串中有空格：
$ IFS=$'\v' read -a foo <<<"$(echo "ABCDE 123456" | sed 's/./&\v/g')"
$ declare -p foo
declare -a foo='([0]="A" [1]="B" [2]="C" [3]="D" [4]="E" [5]=" " [6]="1" [7]="2" [8]="3" [9]="4" [10]="5" [11]="6")'
回答by mkb
If you know it's going to be just two fields, you can skip the extra subprocesses like this:
如果您知道它将只有两个字段，您可以跳过额外的子流程，如下所示：
var1=${STR%-*}
var2=${STR#*-}
What does this do? ${STR%-*}deletes the shortest substring of $STRthat matches the pattern -*starting from the end of the string. ${STR#*-}does the same, but with the *-pattern and starting from the beginning of the string. They each have counterparts %%and ##which find the longestanchored pattern match. If anyone has a helpful mnemonic to remember which does which, let me know! I always have to try both to remember.
这有什么作用？从字符串的末尾开始${STR%-*}删除$STR与模式匹配的最短子-*字符串。${STR#*-}做同样的事情，但使用*-模式并从字符串的开头开始。它们每个都有对应物%%，##并找到最长的锚定模式匹配。如果有人有一个有用的助记符来记住哪个是哪个，请告诉我！我总是必须尝试两者才能记住。
回答by tripleee
Sounds like a job for setwith a custom IFS.
听起来像是set定制的工作IFS。
IFS=-
set $STR
var1=
var2=
(You will want to do this in a function with a local IFSso you don't mess up other parts of your script where you require IFSto be what you expect.)
（您将希望在带有 a 的函数中执行此操作，local IFS这样您就不会弄乱脚本的其他部分，您需要IFS满足您的期望。）
回答by anubhava
Using bash regex capabilities:
使用 bash 正则表达式功能：
re="^([^-]+)-(.*)$"
[[ "ABCDE-123456" =~ $re ]] && var1="${BASH_REMATCH[1]}" && var2="${BASH_REMATCH[2]}"
echo $var1
echo $var2
OUTPUT
输出
ABCDE
123456
