Exporting variables between shell script

Question

I have two independently running scripts. first one lets say script A calculates some values. And i want to echo these values from other script called B. These scripts will not call each other. I have used export keyword but didn't work. How can i do this?

Answer 1

If I understood the requirement then can't both scripts be simply executed in same sub shell, independently but without calling each other or without any need of an external file or pipe like this:

Let's say this is your script1.sh

#!/bin/bash
# script1.sh: do something and finally
export x=19

And here us your script2.sh

#!/bin/bash
# script2.sh: read value of $x here
echo "x="${x}

Just call them in same sub-shell like this

(. ./script1.sh && ./script2.sh)

OUTPUT:

x=19

Answer 2

 mkfifo /tmp/channel

 process_a.sh > /tmp/channel&
 process_b.sh < /tmp/channel&

 wait

Of course you can also just read a single line when you want it.

In bash there are coprocs, which also might be what you want. Random example from this page

# let the output of the coprocess go to stdout
{ coproc mycoproc { awk '{print "foo" $0;fflush()}' ;} >&3 ;} 3>&1
echo bar >&${mycoproc[1]}
foobar

ksh has a similar feature, apparently

Answer 3

Think of each script as a function: function A calculates some value and returns it. It does not know who will call it. Function B takes in some value and echo it. It does not care who produced that value. So, script A is:

#!/bin/sh
# a.sh: Calculate something and return the result
echo 19

and script B is:

#!/bin/sh
# b.sh: Consume the calculated result, which passed in as $1
echo The result is $1

Make them executable:

chmod +x [ab].sh

Now, we can glue them together on the command line:

$ b.sh $(a.sh)
The result is 19

Semantically, b.sh did not call a.sh. You called a.sh and passed its result to b.sh.