How to insert a variable inside bash -c command

Question

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

See: [Difference between single and double quotes in bash](http://stackoverflow.com/q/6697753/3776858) — Cyrus, Nov 29 '17 at 03:10

Luc M · Accepted Answer · 2017-11-29T02:32:56.323

6

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.

edited Nov 29 '17 at 02:32

answered Nov 29 '17 at 02:23

Luc M

2

`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. – codeforester Nov 29 '17 at 05:14

ceilKs · Answer 2 · 2017-11-29T02:38:31.780

-1

you can use bash -c 'FOO=test; echo \$FOO' or export FOO=test;bash -c 'echo $FOO'

edited Nov 29 '17 at 02:38

answered Nov 29 '17 at 02:25

ceilKs

tried that, it won't do, it returns `$FOO` instead. – littleworth Nov 29 '17 at 02:26
export FOO="text" bash -c 'echo $FOO' is the right way, ignore my answer – ceilKs Nov 29 '17 at 02:29
you can also use bash -c '$FOO=test;echo \$FOO', i always do like this – ceilKs Nov 29 '17 at 02:30
@ceilKs update your answer so as to help others that may seek the same assistance – chapskev Nov 29 '17 at 02:34
Sorry but the left part of condition doesn't met the requirement. – Luc M Nov 29 '17 at 02:40

2 Answers2