CODEWITHSUNDEEP
bash
agemO
agemO

Reputation: 195

Prevent parameter propagation

When I call a script inside another one, the parameters I give to the first one are automatically propagate to the second one.

a.sh :

echo "a running"
source b.sh blablabla
source b.sh

b.sh :

echo "b running"
echo $1

Which gives :

$source a.sh hello
a running
b running
blablabla
b running
hello

EDIT :

set ""
echo "a running"
source b.sh blablabla
source b.sh

Can be a solution since set "" set the first parameter to an empty string

Upvotes: 0

Views: 154

Answers (1)

Josh Jolly
Josh Jolly

Reputation: 11796

When you use source, everything inside b.sh is being read and executed as if it were part of a.sh - so it has access to the positional parameters passed to a.sh. What are you trying to accomplish here - is it actually necessary to use source? You can avoid this behaviour by running the script instead of sourcing it:

./b.sh

Or:

bash b.sh

Upvotes: 4

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