Printing arrays are not working in shell script

Question

I am kind of new in shell scripting and trying to learn arrays. I declared array value but when I am trying to print that array it is giving me an error(bad substitution). I am pasting the code below, please suggest to me what is wrong here-

➜  ~ cat test.sh
#!/bin/bash

array=['foo','bar','a','b']
echo 1
echo "${array[0]}"
➜  ~ sh test.sh
1
test.sh: 5: Bad substitution

Thanks in advance.

declare array using '()' **array=('foo' 'bar')** – Volzy May 03 '22 at 17:10 — Volzy, May 03 '22 at 17:10

Arkadiusz Drabczyk · Accepted Answer · 2022-05-03T17:23:33.923

0

Depending on the system you're using sh might be not be Bash and it's not Bash on yours, it can be dash for example. Run your script with Bash:

$ bash arr.sh
1
[foo,bar,a,b]

Or set an executable bit and call the script without providing the name of the interpreter since you already have the shebang:

$ chmod +x test.sh
$ ./test.sh
1
[foo,bar,a,b]

edited May 03 '22 at 17:23

answered May 03 '22 at 17:12

Arkadiusz Drabczyk

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Thank you so much, it was a very silly mistake from my side. I am using zsh. – John The Hacker May 03 '22 at 17:17
You might use zsh as the default _interactive_ shell but the `Bad substitution` message does not come from it, check `zsh test.sh`. – Arkadiusz Drabczyk May 03 '22 at 17:18
Post output of `ls -Al /bin/sh` to find out what's the `sh` mapped to on your system. – Arkadiusz Drabczyk May 03 '22 at 17:18
lrwxrwxrwx 1 root root 4 Aug 20 2021 /bin/sh -> dash – John The Hacker May 03 '22 at 17:19
So it's dash. You'd get the same error if you called test.sh with dash: `dash test.sh` – Arkadiusz Drabczyk May 03 '22 at 17:20
Thank you for pointing out, it was a good learning today. All thanks to you – John The Hacker May 03 '22 at 17:21

Printing arrays are not working in shell script

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