• Uncategorized

About linux : Command-Not-Found-when-assigning-an-integer-variable-from-another-script

Question Detail

To illustrate my issue and questions consider two scripts.


echo $test



When I run Script1.sh from the command line directly via ./Script1.sh it produces the desired output.


However when I run Script2.sh which calls Script1.sh, instead I get the following error message:

Script2.sh: line 1: 1: command not found

It thinks that the integer 1 is a command for some odd reason?

I found a cheap work around which is to set a variable from the first script and pass it to the second script as a parameter but I would like to understand why this is failing.

Also, when I changed my variables value from 1 to “some text”, I would expect that my echo would print “some text” to the screen. But when I call Script2.sh (which ultimately just calls Script1.sh) it doesn’t print the echo to the screen. Why is that, and how do I resolve it so I can see the echo output the same as if I would have called Script1.sh directly?

Any advice would be greatly appreciated.

Question Answer

You’re calling it wrong by using command substitution. Just use source or a subshell if you want it:

. ./Script1.sh
( ./Script1.sh )

The second one is a bit redundant since it will always run in another shell anyway, so this would make more sense instead:

( . ./Script1.sh )

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.