In a Python script, is there any way to tell if the interpreter is in interactive mode? This would be useful so that, for instance, when you run an interactive Python sessio
Use sys.flags:
if sys.flags.interactive:
#interactive
else:
#not interactive
The following works both with and without the -i switch:
#!/usr/bin/python
import sys
# Set the interpreter bool
try:
if sys.ps1: interpreter = True
except AttributeError:
interpreter = False
if sys.flags.interactive: interpreter = True
# Use the interpreter bool
if interpreter: print 'We are in the Interpreter'
else: print 'We are running from the command line'
__main__.__file__
doesn't exist in the interactive interpreter:
import __main__ as main
print hasattr(main, '__file__')
This also goes for code run via python -c
, but not python -m
.
From TFM: If no interface option is given, -i is implied, sys.argv[0] is an empty string ("") and the current directory will be added to the start of sys.path.
If the user invoked the interpreter with python
and no arguments, as you mentioned, you could test this with if sys.argv[0] == ''
. This also returns true if started with python -i
, but according to the docs, they're functionally the same.
Here's something that would work. Put the following code snippet in a file, and assign the path to that file to the PYTHONSTARTUP
environment variable.
__pythonIsInteractive__ = None
And then you can use
if __name__=="__main__":
#do stuff
elif '__pythonIsInteractive__' in globals():
#do other stuff
else:
exit()
http://docs.python.org/tutorial/interpreter.html#the-interactive-startup-file
I compared all the methods I found and made a table of results. The best one seems to be this:
hasattr(sys, 'ps1')
If anyone has other scenarios that might differ, comment and I'll add it