Let's say I declare two variables key and value like this in shell script
$ key=key1
$ value=value1
now I want to create a variable of name key1 and assign it value value1. What I tried is
$ export ${key}
$ export ${value}
$ $($key=$value)
output: key1=value1: command not found
I don't know how to do this.
Use eval $key=$value instead of $($key=$value): The shell substitutes variable values first, and then the $(...) substitution. As there is not command with that name the shell shows command not found in STDERR. Tell the shell to evaluate the substitution result as a regular shell command again by using eval. Later you can export the result with export $key. Using the -x flag for the shell provides a good insight of what happens:
$ key=key1 $ value=value1 $ set -x $ $key=$value + key1=value1 sh: key1=value1: not found [No such file or directory] $ $($key=$value) + key1=value1 sh: key1=value1: not found [No such file or directory] $ eval $key=$value + eval key1=value1 + key1=value1 $ echo $key1 + echo value1 value1 $ export $key + export key1
Also, be careful when dealing with variables this way: Whitespaces in shell variables can have unexpected results in this kind of constructions.
bashsupports${!key}to do indirect variable referencing. If you're not in charge of the contents of the variables, be very careful of theevalmethod.key="$(rm -fr / 2>/dev/null & echo Hi)"is not going to lead to happiness.