Why Parameter Expansion is not working? [duplicate] - sh

Question

#!/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.

Answer 1

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.

Answer 2

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

Answer 3

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)

Answer 4

Your script syntax is valid bash and good.
Possible causes for the failure:
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.
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.

Answer 5

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

Answer 6

Also, make sure you don't have an empty string for the first line of your script.
i.e. make sure #!/bin/bash is the very first line of your script.

Answer 7

Not relevant to your example, but you can also get the Bad substitution error in Bash for any substitution syntax that Bash does not recognize. This could be:
Stray whitespace. E.g. bash -c '${x }'
A typo. E.g. bash -c '${x;-}'
A feature that was added in a later Bash version. E.g. bash -c '${x#Q}' before Bash 4.4.
If you have multiple substitutions in the same expression, Bash may not be very helpful in pinpointing the problematic expression. E.g.:
$ bash -c '"${x } multiline string
$y"'
bash: line 1: ${x } multiline string
$y: bad substitution

Answer 8

Both - bash or dash - work, but the syntax needs to be:
FILENAME=/my/complex/path/name.ext
NEWNAME=${FILENAME%ext}new

Answer 9

I was adding a dollar sign twice in an expression with curly braces in bash:
cp -r $PROJECT_NAME ${$PROJECT_NAME}2
instead of
cp -r $PROJECT_NAME ${PROJECT_NAME}2

Answer 10

I have found that this issue is either caused by the marked answer or you have a line or space before the bash declaration

Answer 11

Looks like "+x" causes problems:
root#raspi1:~# cat > /tmp/btest
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
root#raspi1:~# chmod +x /tmp/btest
root#raspi1:~# /tmp/btest
root#raspi1:~# sh -x /tmp/btest
+ jobname=job_201312161447_0003
/tmp/btest: 4: /tmp/btest: Bad substitution

Answer 12

in my case (under ubuntu 18.04), I have mixed $( ${} ) that works fine:
BACKUPED_NB=$(ls ${HOST_BACKUP_DIR}*${CONTAINER_NAME}.backup.sql.gz | wc --lines)
full example here.

Answer 13

I used #!bin/bash as well tried all approaches like no line before or after #!bin/bash.
Then also tried using +x but still didn't work.
Finally i tried running the script ./script.sh it worked fine.
#!/bin/bash
jobname="job_201312161447_0003"
jobname_post=${jobname:17}
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# sh jaru.sh
jaru.sh: 3: jaru.sh: Bad substitution
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# ./jaru.sh
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts#