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"

Note that printf will iterate if you provide more values than placeholders. Your for loop can be written like this: bash -c 'arr=(1 2 3 4 5); printf "%s\n" "${arr[@]}"' — glenn jackman, Sep 10 '15 at 13:29

score 6 · Answer 1 · edited Apr 13 '17 at 12:36

6

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

edited Apr 13 '17 at 12:36

Community

1

answered Sep 10 '15 at 10:46

cuonglm

153,898

I'd also recommend to attach shebang https://en.wikipedia.org/wiki/Shebang_%28Unix%29 at the beggining of EACH script. – Mateusz Pacek Sep 10 '15 at 10:49
2

@MateuszPacek: The OP calling sh -c inline script instead of creating a shell script. the shebang didn't help in this case. – cuonglm Sep 10 '15 at 10:51
@ccuonglm yes, that's true. But when you got script in file it's easier to exec it inline if you know what kind of shell implementation you used to use. – Mateusz Pacek Sep 10 '15 at 10:55
this works with single quote sh -c 'arr=(1 2 3 4 5);for var in ${arr[@]};do echo $var;done' – c4f4t0r Sep 10 '15 at 12:57
2

@c4f4t0r this works only if your sh isn't really sh but is actually bash, zsh, or ksh in poor disguise. You shouldn't really use bash features with the shell invoked as sh as one day it may break (consider Debian with sh a symlink to dash rather than bash). – Chris Davies Sep 10 '15 at 13:05
yes, I'm using centos and sh is symlink to bash – c4f4t0r Sep 10 '15 at 13:21
1

@c4f4t0r, still, avoid bad practices: use the most precise interpreter for your code. – glenn jackman Sep 10 '15 at 13:28

score 2 · Answer 2 · edited Apr 13 '17 at 12:36

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'

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

2 Answers2