I can do the following:
$ FOO="text"
$ echo $FOO
$ text
But how can I wrap it inside bash -c construct? I tried this but failed:
$ FOO="text"
$ bash -c 'echo $FOO'
$ # return nothing
The reason I ask this because I need to execute another 3rd party code that need to be wrapped inside bash -c
Try
$ export FOO="text"
$ bash -c 'echo $FOO'
export command is used to export a variable or function to the environment of all the child processes running in the current shell.
The "bash" command starts a child process where its parent is your current bash session.
To define a variable in parent process and use it in child process, you have to export it.
FOO=text bash -c 'echo $FOO' would work as well - in this case, the environment variable FOO is set just in the context of the process (bash) being created.you can use bash -c 'FOO=test; echo \$FOO'
or export FOO=test;bash -c 'echo $FOO'
$FOO instead.
FOO="text"; bash -c "echo $FOO"