Replacing substring for another string in bash script

Question

I'm creating shell script in Ubuntu 20.04 VM, so I'm testing in bash the following:

echo ${"/mnt/raid1/"//\//}

What I need to achieve is to get the folder's route as valid variable name, something like 'mntraid1'. However previous code throws bash error - bad substitution.

Any orientation appreciated.

In particular, sh isn't guaranteed to support ${var//find/subst} at all. — Charles Duffy
– Charles Duffy, Commented Apr 24, 2022 at 19:08
My actual goal is to create a shell script but since it wasn't working I'm testing it on bash, I'm sorry if they are different questions, I actually don't know. — MatiasGSP
– MatiasGSP, Commented Apr 24, 2022 at 19:11
If you do want to use this feature, be sure you're writing a bash script instead of a sh script. They're both "shell scripts", but bash guarantees a larger set of features. — Charles Duffy
– Charles Duffy, Commented Apr 24, 2022 at 19:11
(Moreover, if you're dynamically generating variable names, it's much easier to work with them using other bash extensions -- with baseline POSIX sh you're stuck with eval, and there's a lot of security impact when it's used carelessly; whereas bash gives you indirect references so you can use "variable variables" safely). — Charles Duffy
– Charles Duffy, Commented Apr 24, 2022 at 19:13

Charles Duffy · Accepted Answer · 2022-04-24 19:53:51Z

1

Only a variable name can be in the first position of this parameter expansion. Thus, you must assign your string to a variable before running a substitution on it.

#!/usr/bin/env bash
#              ^^^^- this makes your script a bash script

# for extra paranoia, one can add an explicit version check
# otherwise, a user running 'sh yourscript' can override the shebang above
case $BASH_VERSION in '') echo "This script requires bash" >&2; exit 1;; esac

dev=/mnt/raid1
var=${dev//'/'/}
echo "Variable name corresponding to $dev is $var"

Note that this is an extension not guaranteed by the POSIX sh standard. It is thus unsafe to use in any script with a #!/bin/sh shebang (which only guarantees functionality required by that standard).

edited Apr 24, 2022 at 19:53

answered Apr 24, 2022 at 19:08

Charles Duffy

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1 Comment

MatiasGSP Over a year ago

Thanks for your answer, I think my problem has to do with not being able make my script a bash one, because the code provided by you is working in a VM with another OS. When I add !/usr/bin/env bash, it throws not found error.

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