sh -c not get the right output when execute shell script

Question

I loop to get data from array in shell ,
It works when I execute it in a shell file, with this content:

arr=(1 2 3 4 5)  
for var in ${arr[@]};  
do  
    echo $var  
done

But there isn't any output when I use sh -c like below:

sh -c "arr=(1 2 3 4 5);for var in ${arr[@]};do echo $var;done"

Answer 1

Your problem is using sh -c "...", see @Gilles's answer for more details.

Further more, sh (refer to POSIX sh) does not support array (strictly speaking, it has only one array, $@), you need to call other shells on your system, which support array like bash, zsh or ksh.

bash -c 'arr=(1 2 3 4 5);for var in "${arr[@]}";do echo "$var"; done'

Also note that you have a mistake when leaving ${arr[@]} un-quote, actually you need for var in "${arr[@]}" instead. Invoking variable without quotes calling split+glob and is source of many security implication.

To play with POSIX sh, you can use $@:

set -- 1 2 3 4 5
for var do
  printf '%s\n' "$var"
done

Answer 2

The problem is that you used the wrong type of quotes. Shells interpolate variables inside double quotes, so the variable references were interpolated by the shell in which you ran sh -c "…". Assuming that this shell didn't have variables named arr or var defined, the argument to sh -c was

arr=(1 2 3 4 5);for var in ;do echo ;done

If sh is a shell that supports arrays, this is perfectly valid code, that happens to do nothing.

To pass a literal string to a command, use single quotes. All characters inside single quotes are interpreted literally, so you can use every character except a single quote in the string. (If you need a single quote, escape it as the four-character sequence '\''.)

sh -c 'arr=(1 2 3 4 5);for var in ${arr[@]};do echo $var;done'

Beware that not all implementations of sh support arrays. If you need arrays, you should explicitly invoke a shell that supports them.

ksh -c 'arr=(1 2 3 4 5);for var in ${arr[@]};do echo $var;done'

or

bash -c 'arr=(1 2 3 4 5);for var in ${arr[@]};do echo $var;done'