BASH: Variable Passing Between Two Scripts

bits45 · 12-20-2014, 08:36 AM

I'm not sure I can do this.

I have script-A execute a bunch of stuff. To make it shorter I broke some of out of of the "stuff" to another script, called script-B.

In script-B I collect info from user and I export the variables I create.

Script-A is not picking up those exported variables as Script-A continues to do perform more "stuff" after script-B finishes. Using bash's "set -u" option show's that the variables are unset in the eyes of script-A.

How do I pass script-B's variables back to script-A if the export VARIABLE entries do not pass back the values? I do source other values in a common file, so all that works fine.

I really don't want to write the values into files only to have script-A suck them back in, that seems lame, Yes?

So, looks something like this.

script-A ---> executes script-B ---> script-A continues looking for B's variables.

Thanks!

Ser Olmy · 12-20-2014, 08:50 AM

When you call script B from script A, script B runs in a separate instance of the shell. This instance receives a copy of the current environment, hence all variables from script A will be available to script B. However, once script B ends, control is returned to the shell instance running script A, and any changes script B may have made to its environment is lost.

Solution: Rather than calling script B from script A, use the "source" command/function:

Code:

[... this is script A ...]

. ./path/to/scriptb

[... rest of script A ...]

Note the dot before the path; that's the "source" command. It cause the entire contents of script B to be "imported" into script A, and thus it will run in the same shell instance as script A.

bits45 · 12-20-2014, 11:01 AM

Well heck, that must of been a really stupid question to ask.

Naturally that worked great. I guess I couldn't see the forest thru the trees.

Thanks for the quick reply.

Worked great.

take care