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#!/bin/bash



jobname="job_201312161447_0003"

jobname_pre=${jobname:0:16}

jobname_post=${jobname:17}

This bash script gives me Bad substitution error on Ubuntu. Any help will be highly appreciated.

The default shell (/bin/sh) under Ubuntu points to dash, not bash.

me@pc:~$ readlink -f $(which sh)

/bin/dash

So if you chmod +x your_script_file.sh and then run it with ./your_script_file.sh, or if you run it with bash your_script_file.sh, it should work fine.

Running it with sh your_script_file.sh will not work because the hashbang line will be ignored and the script will be interpreted by dash, which does not support that string substitution syntax.

I had the same problem. Make sure your script didnt have

#!/bin/sh 

at the top of your script. Instead, you should add

#!/bin/bash

Your script syntax is valid bash and good.

Possible causes for the failure:

1. Your bash is not really bash but ksh or some other shell which doesn't understand bash's parameter substitution. Because your script looks fine and works with bash.
   Do ls -l /bin/bash and check it's really bash and not sym-linked to some other shell.




2. If you do have bash on your system, then you may be executing your script the wrong way like: ksh script.sh or sh script.sh (and your default shell is not bash). Since you have proper shebang, if you have bash ./script.sh or bash ./script.sh should be fine.

For others that arrive here, this exact message will also appear when using the env variable syntax for commands, for example ${which sh} instead of the correct $(which sh)

Try running the script explicitly using bash command rather than just executing it as executable.