获取Python中当前脚本的名称

I’m trying to get the name of the Python script that is currently running.

I have a script called foo.py and I’d like to do something like this in order to get the script name:

print Scriptname

You can use __file__ to get the name of the current file. When used in the main module, this is the name of the script that was originally invoked.

If you want to omit the directory part (which might be present), you can use os.path.basename(__file__).

import sys
print sys.argv[0]

This will print foo.py for python foo.py, dir/foo.py for python dir/foo.py, etc. It’s the first argument to python. (Note that after py2exe it would be foo.exe.)

For completeness’ sake, I thought it would be worthwhile summarizing the various possible outcomes and supplying references for the exact behaviour of each:

• __file__ is the currently executing file, as detailed in the official documentation:

  __file__ is the pathname of the file from which the module was loaded, if it was loaded from a file. The __file__ attribute may be missing for certain types of modules, such as C modules that are statically linked into the interpreter; for extension modules loaded dynamically from a shared library, it is the pathname of the shared library file.

  From Python3.4 onwards, per issue 18416, __file__ is always an absolute path, unless the currently executing file is a script that has been executed directly (not via the interpreter with the -m command line option) using a relative path.

• __main__.__file__ (requires importing __main__) simply accesses the aforementioned __file__ attribute of the main module, e.g. of the script that was invoked from the command line.

• sys.argv[0] (requires importing sys) is the script name that was invoked from the command line, and might be an absolute path, as detailed in the official documentation:

  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.

  As mentioned in another answer to this question, Python scripts that were converted into stand-alone executable programs via tools such as py2exe or PyInstaller might not display the desired result when using this approach (i.e. sys.argv[0] would hold the name of the executable rather than the name of the main Python file within that executable).

• If none of the aforementioned options seem to work, probably due to an irregular import operation, the inspect module might prove useful. In particular, invoking inspect.getfile(...) on inspect.currentframe() could work, although the latter would return None when running in an implementation without Python stack frame.

Handling symbolic links

If the current script is a symbolic link, then all of the above would return the path of the symbolic link rather than the path of the real file and os.path.realpath(...) should be invoked in order to extract the latter.

Further manipulations that extract the actual file name

may be invoked on any of the above in order to extract the actual file name and may be invoked on the actual file name in order to truncate its suffix, as in os.path.splitext(os.path.basename(...)).

From Python 3.4 onwards, per PEP 428, the PurePath class of the pathlib module may be used as well on any of the above. Specifically, pathlib.PurePath(...).name extracts the actual file name and pathlib.PurePath(...).stem extracts the actual file name without its suffix.

Note that __file__ will give the file where this code resides, which can be imported and different from the main file being interpreted. To get the main file, the special __main__ module can be used:

import __main__ as main
print(main.__file__)

Note that __main__.__file__ works in Python 2.7 but not in 3.2, so use the import-as syntax as above to make it portable.

The Above answers are good . But I found this method more efficient using above results.
This results in actual script file name not a path.

import sys    
import os    
file_name =  os.path.basename(sys.argv[0])

For modern Python versions (3.4+), Path(__file__).name should be more idiomatic. Also, Path(__file__).stem gives you the script name without the .py extension.

Try this:

print __file__

Note: If you are using Python 3+, then you should use the print() function instead

Assuming that the filename is foo.py, the below snippet

import sys
print sys.argv[0][:-3]

or

import sys
print sys.argv[0][::-1][3:][::-1]

As for other extentions with more characters, for example the filename foo.pypy

import sys
print sys.argv[0].split('.')[0]

If you want to extract from an absolute path

import sys
print sys.argv[0].split('/')[-1].split('.')[0]

will output foo

The first argument in sys will be the current file name so this will work

   import sys
   print sys.argv[0] # will print the file name

If you’re doing an unusual import (e.g., it’s an options file), try:

import inspect
print (inspect.getfile(inspect.currentframe()))

Note that this will return the absolute path to the file.

we can try this to get current script name without extension.

import os

script_name = os.path.splitext(os.path.basename(__file__))[0]

Since the OP asked for the name of the current script file I would prefer

import os
os.path.split(sys.argv[0])[1]

My fast dirty solution:

__file__.split('/')[-1:][0]

will give you an absolute path (relpath() available as well).

sys.argv[-1] will give you a relative path.

all that answers are great, but have some problems You might not see at the first glance.

lets define what we want – we want the name of the script that was executed, not the name of the current module – so __file__ will only work if it is used in the executed script, not in an imported module. sys.argv is also questionable – what if your program was called by pytest ? or pydoc runner ? or if it was called by uwsgi ?

and – there is a third method of getting the script name, I havent seen in the answers – You can inspect the stack.

Another problem is, that You (or some other program) can tamper around with sys.argv and __main__.__file__ – it might be present, it might be not. It might be valid, or not. At least You can check if the script (the desired result) exists !

my library bitranox/lib_programname at github does exactly that :

• check if __main__ is present
• check if __main__.__file__ is present
• does give __main__.__file__ a valid result (does that script exist ?)
• if not: check sys.argv:
• is there pytest, docrunner, etc in the sys.argv ? –> if yes, ignore that
• can we get a valid result here ?
• if not: inspect the stack and get the result from there possibly
• if also the stack does not give a valid result, then throw an Exception.

by that way, my solution is working so far with setup.py test, uwsgi, pytest, pycharm pytest , pycharm docrunner (doctest), dreampie, eclipse

there is also a nice blog article about that problem from Dough Hellman, “Determining the Name of a Process from Python”

As of Python 3.5 you can simply do:

from pathlib import Path
Path(__file__).stem

See more here: https://docs.python.org/3.5/library/pathlib.html#pathlib.PurePath.stem

For example, I have a file under my user directory named test.py with this inside:

from pathlib import Path

print(Path(__file__).stem)
print(__file__)

running this outputs:

>>> python3.6 test.py
test
test.py