问题
Quoting from docs.python.org:
"sys.argv The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string."
Am I missing something, or sys.argv[0] always returns the script name, and to get '-c' I'd have to use sys.argv[1]?
I'm testing with Python 3.2 on GNU/Linux.
回答1:
No, if you invoke Python with -c to run commands from the command line, your sys.argv[0] will be -c:
C:\Python27>python.exe -c "import sys; print sys.argv[0]"
-c
回答2:
When Python is invoked as python script.py then sys.argv[0] == 'script.py'. When you invoke python -c 'import sys; print sys.argv' then sys.argv[0] == '-c' indicating the script body was passed as a string on the command line.
回答3:
python -c executes a command passed on the command line, rather than a script from a file. sys.argv[0] will be set to "-c".
If you run a script with a -c flag, then yes, sys.argv[1] will be set to "-c" and sys.argv[0] will be set to the name of the script.
来源：https://stackoverflow.com/questions/5222408/python-sys-argv0-meaning-in-official-documentation