Return values from one script to another script

Question

I have the following script that will run each script (sequentially) in a directory:

import os directory = [] for dirpath, dirnames, filenames in os.walk("path\to\scripts"): for filename in [f for f in filenames if f.endswith(".py")]: directory.append(os.path.join(dirpath, filename)) for entry in directory: execfile(entry) print x

my scripts look like this:

def script1(): x = "script 1 ran" return x script1()

When print x is called, it says x is not defined. I'm just curious if there is a way to return values so that the parent script can access the data.

stackoverflow.com/questions/4807680/… check this out.

John Riselvato
– John Riselvato

2011-09-16 18:10:34 +00:00
Commented Sep 16, 2011 at 18:10 — John Riselvato
– John Riselvato, Commented Sep 16, 2011 at 18:10

S.Lott · Accepted Answer · 2011-09-16 18:13:17Z

I'm just curious if there is a way to return values so that the parent script can access the data.

This is why you define functions and return values.

Script 1 should include a function.

def main(): all the various bits of script 1 except the import return x if __name__ == "__main__": x= main() print( x )

Works the same as yours, but now can be used elsewhere

Script 2 does this.

import script1 print script1.main()

That's the way one script uses another.

Staven · Accepted Answer · 2011-09-16 18:41:04Z

You can use the locals argument of execfile(). Write the scripts like this:

def run_script(): ret_value = 2 return ret_value script_ret = run_script()

And in your main script:

script_locals = dict() execfile("path/to/script", dict(), script_locals) print(script_locals["script_ret"])

Tom Zych · Accepted Answer · 2011-09-16 18:12:22Z

x is local to the script1 function, so it's not visible in the outer scope. If you put the code inside script1 at the top level of the file, it should work.

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