Variable substitution with an exclamation mark in bash
I have the following lines in my .cfg bash script file
variable-substitution
${!FOO} and zsh
${!FOO} performs a double substitution in bash, meaning it takes the (string) value of FOO and uses it as a variable name.
zsh doesn’t support this feature.
Can command substitution be nested in variable substitution?
I would like to use variable substitution on a particular string that I access via a command. For example, if I copy something into my clipboard, I can access it like this.
Is it possible to print the content of the content of a variable with shell script? (indirect referencing)
Let’s suppose I’ve declared the following variables:
Behaviour of bash command substitution with command from string in variable
There are seven kinds of expansion … The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and filename expansion
Bash variable substitution of variable followed by underscore
The variable BUILDNUMBER is set to value 230. I expect 230_ to be printed for the command echo $BUILDNUMBER_ but the output is empty as shown below.