Question

a.sh

#! /bin/sh
export x=/usr/local

we can do source ./a in command-line. But I need to do the export through shell script.

b.sh

#! /bin/sh
. ~/a.sh

no error… but $x in command-line will show nothing. So it didn’t get export.

Any idea how to make it work?

---

a.sh

#! /bin/sh
export x=/usr/local
-----------
admin@client: ./a.sh
admin@client: echo $x

admin@client:  <insert ....>

Answer 1

You can’t do an export through a shell script, because a shell script runs in a child shell process, and only children of the child shell would inherit the export.

The reason for using source is to have the current shell execute the commands

It’s very common to place export commands in a file such as .bashrc which a bash will source on startup (or similar files for other shells)

Another idea is that you could create a shell script which generates an export command as it’s output:

shell$ cat > script.sh
#!/bin/sh
echo export foo=bar
^D
chmod u+x script.sh

And then have the current shell execute that output

shell$ `./script.sh`

shell$ echo $foo
bar

shell$ /bin/sh
$ echo $foo
bar

(note above that the invocation of the script is surrounded by backticks, to cause the shell to execute the output of the script)