Passing a pre-defined variable from shell script to shell script in Linux

Question

This is my first time using a shell script.

I have one "Mother" script mainscript.sh, in which I define the variable patientid.

patientid=`basename $folder`

Later on in the same script, I wish to execute a separate script example.sh while passing the variable patientid into it. That script already has the variable labeled as "$patientid" in it. Looking at mainscript.sh below:

./example.sh #I WANT TO PASS THE VARIABLE patientid INTO HERE!

I know this is easy-peasy for y'all. Any help is appreciated! Thanks.

Does example.sh only refer to $patientid or does it set it somewhere, like in patientid=somevalue? — Jens
– Jens, Commented Jan 27, 2015 at 21:27

Wintermute · Accepted Answer · 2015-01-27 21:17:15Z

1

Before calling example.sh, mark patientid for export to the environment of child processes (such as the shell that will run example.sh):

export patientid

answered Jan 27, 2015 at 21:17

Wintermute

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Jens · Accepted Answer · 2015-01-27 21:29:17Z

1

You can run any shell script (in fact, even binary programs) with a set of variables predefined with

 VAR1=value1 VAR2=value2 ... script.sh

e.g.

 patientid=$patientid mainscript.sh

This assumes a Bourne-heritage shell (sh, bash, ash, ksh, zsh, ...)

answered Jan 27, 2015 at 21:29

Jens

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Marat Talipov · Accepted Answer · 2015-01-27 21:11:34Z

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Perhaps, what you need is to pass the parameter through the command line arguments, i.e.:

./example.sh $patientid

in the main script and

patientid=$1

in the example.sh script.

answered Jan 27, 2015 at 21:11

Marat Talipov

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2 Comments

Henri Wathieu Over a year ago

so, having patientid=$1 will automatically define it in example.sh the way it was defined in mainscript.sh? Just by adding $patientid after the execution command?

Marat Talipov Over a year ago

Right. ./example.sh $patientid will execute ./example.sh with $patientid as the command line argument. On the other side, the script example.sh will be able to access that argument as $1 (which is a standard name for the first command line argument) and assign it to patientid by doing patientid=$1. Obviously, this command will overwrite the old content of patientid in example.sh.

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