I want to do something like this
我想做这样的事情
A='123'
B='143'
C='999'
declare -a arr=(A B C)
for i in "{$arr[@]}"
do
echo "@i" "$i"
done
Which should give me the output of
哪个应该给我输出
A 123
B 143
C 999
But instead I receive the variable names, not the value in the output (I just see "A @i" in the output...
但我得到的是变量名,而不是输出中的值(我只是在输出中看到“A @i”...
2 个解决方案
#1
2
If you want to store the variable names in the loop, rather than copy their values, then you can use the following:
如果要将变量名存储在循环中,而不是复制它们的值,则可以使用以下命令:
for i in "${arr[@]}"; do
echo "${!i}"
done
This means that the value of i is taken as a name of a variable, so you end up echoing $A, $B and $C in the loop.
这意味着i的值被视为变量的名称,因此您最终会在循环中回显$ A,$ B和$ C.
Of course, this means that you can print the variable name at the same time, e.g. by using:
当然,这意味着您可以同时打印变量名称,例如通过使用:
echo "$i: ${!i}"
It's not exactly the same, but you may also be interested in using an associative array:
它不完全相同,但您可能也对使用关联数组感兴趣:
declare -A assoc_arr=( [A]='123' [B]='143' [C]='999' )
for key in "${!assoc_arr[@]}"; do
echo "$key: ${assoc_arr[$key]}"
done
#2
1
I suggest to add $:
我建议添加$:
declare -a arr=("$A" "$B" "$C")
#1
2
If you want to store the variable names in the loop, rather than copy their values, then you can use the following:
如果要将变量名存储在循环中,而不是复制它们的值,则可以使用以下命令:
for i in "${arr[@]}"; do
echo "${!i}"
done
This means that the value of i is taken as a name of a variable, so you end up echoing $A, $B and $C in the loop.
这意味着i的值被视为变量的名称,因此您最终会在循环中回显$ A,$ B和$ C.
Of course, this means that you can print the variable name at the same time, e.g. by using:
当然,这意味着您可以同时打印变量名称,例如通过使用:
echo "$i: ${!i}"
It's not exactly the same, but you may also be interested in using an associative array:
它不完全相同,但您可能也对使用关联数组感兴趣:
declare -A assoc_arr=( [A]='123' [B]='143' [C]='999' )
for key in "${!assoc_arr[@]}"; do
echo "$key: ${assoc_arr[$key]}"
done
#2
1
I suggest to add $:
我建议添加$:
declare -a arr=("$A" "$B" "$C")