bash - Sharing variables between shell scripts -

i have shell script (let's call parent.sh) calls shell script (child.sh), passing argument.

the child script work , sets value in variable called create_sql. want access create_sql variable parent script invoked it.

i calling child script within parent script this:

./child.sh "$dictionary"

and straight afterwards have line:

echo "the resulting create script is: "$create_sql

however, there no value being output, however, in child script doing same thing , variable set.

how can work can read variables created child script?

succinctly, can't have child script set variable in parent script unless like:

. ./child.sh "$dictionary" 

(or in bash, mimicking c shell, source ./child.sh "$dictionary"). reads , executes script in environment of current shell, alter other variable in parent.sh script; there no isolation between scripts. otherwise, child process cannot sanely set environment of parent shell. (if want insanely, can have child shell run debugger, attach parent shell process , set environment way — calling 'insane' being polite).

arguably, best way output child stashed in variable in parent script have child.sh echo value want in $create_sql, , use

create_sql=$(./child.sh "$dictionary") echo "the resulting create script is: $create_sql" 

with no spaces around assignment operator. note echo includes variable inside double quotes; preserve internal spacing (including newlines) in variable's value. written in question, variable flattened space-separated stream of 'words' (sequences of non-spaces), losing internal spacing.