英文:
How do I perform double substitiution of shell variables?
问题
我有以下脚本存储在test.sh中：
#!/bin/sh
key1=foo
key2=bar
key3=baz
for i in 1 2 3; do
echo $i
echo ${key${i}}
done
当我运行它时，我得到以下输出：
1
test.sh: 9: Bad substitution
我做错了什么？
英文:
I have the following script stored in test.sh:
#!/bin/sh
key1=foo
key2=bar
key3=baz
for i in 1 2 3; do
echo $i
echo ${key${i}}
done
When I run it I get the output
1
test.sh: 9: Bad substitution
What am I doing wrong?
答案1
得分: 1
你可以尝试
eval echo '$'key$i
英文:
You can try
eval echo '$'key$i
答案2
得分: 1
这是你在寻找的吗？
#!/bin/sh
key1=foo
key2=bar
key3=baz
for i in 1 2 3; do
echo $i
eval echo $"key$i"
done
输出：
1
foo
2
bar
3
baz
英文:
Is this what you are looking for?
#!/bin/sh
key1=foo
key2=bar
key3=baz
for i in 1 2 3; do
echo $i
eval echo $"key$i"
done
Output:
1
foo
2
bar
3
baz
答案3
得分: 0
这里有这个问题的答案：https://stackoverflow.com/a/917313/10693596
简短回答是在bash中不可能实现。
更新：根据你最终的需求，使用间接扩展可能会起作用：https://stackoverflow.com/a/42066820/10693596
英文:
There is an answer to this question here: https://stackoverflow.com/a/917313/10693596
Short answer is this is not possible in bash.
Update: depending on what you are ultimately after, using indirect expansion might work: https://stackoverflow.com/a/42066820/10693596
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