How to call a script from another script?

This is possible in Python 2 using

execfile("test2.py")

See the documentation for the handling of namespaces, if important in your case.

In Python 3, this is possible using (thanks to @fantastory)

exec(open("test2.py").read())

However, you should consider using a different approach; your idea (from what I can see) doesn’t look very clean.

Why not just import test1? Every python script is a module. A better way would be to have a function e.g. main/run in test1.py, import test1 and run test1.main(). Or you can execute test1.py as a subprocess.

You should not be doing this. Instead, do:

test1.py:

 def print_test():
      print "I am a test"
      print "see! I do nothing productive."

service.py

#near the top
from test1 import print_test
#lots of stuff here
print_test()

Use import test1 for the 1st use – it will execute the script. For later invocations, treat the script as an imported module, and call the reload(test1) method.

When reload(module) is executed:

• Python modules’ code is recompiled and the module-level code reexecuted, defining a new set of objects which are bound to names in the module’s dictionary. The init function of extension modules is not called

A simple check of sys.modules can be used to invoke the appropriate action. To keep referring to the script name as a string ('test1'), use the ‘import()’ builtin.

import sys
if sys.modules.has_key['test1']:
    reload(sys.modules['test1'])
else:
    __import__('test1')

If you want test1.py to remain executable with the same functionality as when it’s called inside service.py, then do something like:

test1.py

def main():
    print "I am a test"
    print "see! I do nothing productive."

if __name__ == "__main__":
    main()

service.py

import test1
# lots of stuff here
test1.main() # do whatever is in test1.py

Another way:

File test1.py:

print "test1.py"

File service.py:

import subprocess

subprocess.call("test1.py", shell=True)

The advantage to this method is that you don’t have to edit an existing Python script to put all its code into a subroutine.

Documentation: Python 2, Python 3

import os

os.system("python myOtherScript.py arg1 arg2 arg3")  

Using os you can make calls directly to your terminal. If you want to be even more specific you can concatenate your input string with local variables, ie.

command = 'python myOtherScript.py ' + sys.argv[1] + ' ' + sys.argv[2]
os.system(command)

This is an example with subprocess library:

import subprocess

python_version = '3'
path_to_run = './'
py_name = '__main__.py'

# args = [f"python{python_version}", f"{path_to_run}{py_name}"]  # works in python3
args = ["python{}".format(python_version), "{}{}".format(path_to_run, py_name)]

res = subprocess.Popen(args, stdout=subprocess.PIPE)
output, error_ = res.communicate()

if not error_:
    print(output)
else:
    print(error_)

I prefer runpy:

#!/usr/bin/env python
# coding: utf-8

import runpy

runpy.run_path(path_name='script-01.py')
runpy.run_path(path_name='script-02.py')
runpy.run_path(path_name='script-03.py')

As it’s already mentioned, runpy is a nice way to run other scripts or modules from current script.

By the way, it’s quite common for a tracer or debugger to do this, and under such circumstances methods like importing the file directly or running the file in a subprocess usually do not work.

It also needs attention to use exec to run the code. You have to provide proper run_globals to avoid import error or some other issues. Refer to runpy._run_code for details.

This process is somewhat un-orthodox, but would work across all python versions,

Suppose you want to execute a script named ‘recommend.py’ inside an ‘if’ condition, then use,

if condition:
       import recommend

The technique is different, but works!

Add this to your python script.

import os
os.system("exec /path/to/another/script")

This executes that command as if it were typed into the shell.

An example to do it using subprocess.

from subprocess import run

import sys

run([sys.executable, 'fullpathofyourfile.py'])

According to the given example, this is the best way:

# test1.py

def foo():
    print("hellow")

# test2.py
from test1 import foo # might be different if in different folder.
foo()

But according to the title, using os.startfile("path") is the best way as its small and it works. This would execute the file specified. My python version is 3.x +.

I found runpy standard library most convenient. Why? You have to consider case when error raised in test1.py script, and with runpy you are able to handle this in service.py code. Both traceback text (to write error in log file for future investigation) and error object (to handle error depends on its type): when with subprocess library I wasn’t able to promote error object from test1.py to service.py, only traceback output.
Also, comparing to "import test1.py as a module" solution, runpy is better cause you have no need to wrap code of test1.py into def main(): function.

Piece of code as example, with traceback module to catch last error text:

import traceback
import runpy #https://www.tutorialspoint.com/locating-and-executing-python-modules-runpy

from datetime import datetime


try:
    runpy.run_path("./E4P_PPP_2.py")
except Exception as e:
    print("Error occurred during execution at " + str(datetime.now().date()) + " {}".format(datetime.now().time()))
    print(traceback.format_exc())
    print(e)