How to call a function from other module

script code at indent level 0 (function1()
function2() in script2 in this case) is executed at import time.

so script2 should only be:
from sikuli import *

def function1(): #
    popup("hello, I am script2, function1")

def function2():
    popup("hello, I am script2, function2")

... and script1 can either be
import Script2
popup("hello, I am script1")
Script2.function2()

... or
... and script1 can either be
from Script2 import *
popup("hello, I am script1")
function2()

... in the second case it might come to name clashes.

Hi RaiMan,

Thanks for the reply.
If I changed the Script2 as per the suggestion, I will not get any output. Script2 will act as a library file....

For me, Script2 also should work, if I run Script2 alone individually. Correct me if am Wrong

I always recommend, to strictly differentiate between library scripts (no executable code (indent level 0)) and execution scripts.

Library scripts contain classes and functions, that are used elsewhere more than once.

An execution script represents a workflow, where one or more classes/functions are used from imported library scripts.

As said - my recommendation based on experience.

But if you really want to: level 0 code in library scripts should be guarded with an if-statement, that only gets true if the library script is run as main script.

The most often used snippet for this case is
if __name__ == "__main__":
    # here your code, that only runs, if the module is a main script (not imported)

Thanks RaiMan !