__file__ 不是您要找的.​​strong> 不要使用意外的副作用>

sys.argv[0] 是总是脚本的路径(如果实际上已经调用了一个脚本)——参见 http://docs.python.org/library/sys.html#sys.argv

__file__ 是当前执行文件(脚本或模块)的路径.如果它是从脚本访问的，那么这意外与脚本相同！如果你想把一些有用的东西，比如定位相对于脚本位置的资源文件放到一个库中，那么你必须使用 sys.argv[0].

示例:

C:\junk\so>输入\junk\so\scriptpath\script1.py导入系统，操作系统打印脚本:sys.argv[0] 是"，repr(sys.argv[0])打印脚本:__file__ 是"，repr(__file__)打印脚本:cwd 是"，repr(os.getcwd())导入 whereutilswhereutils.show_where()C:\junk\so>type \python26\lib\site-packages\whereutils.py导入系统，操作系统def show_where():打印show_where:sys.argv[0] 是"，repr(sys.argv[0])打印 "show_where: __file__ is", repr(__file__)打印 "show_where: cwd is", repr(os.getcwd())C:\junk\so>\python26\python scriptpath\script1.py脚本:sys.argv[0] 是 'scriptpath\\script1.py'脚本: __file__ 是 'scriptpath\\script1.py'脚本:cwd 是 'C:\\junk\\so'show_where: sys.argv[0] 是 'scriptpath\\script1.py'show_where: __file__ 是 'C:\\python26\\lib\\site-packages\\whereutils.pyc'show_where: cwd 是 'C:\\junk\\so'

How can I find the full path to the currently running Python script? That is to say, what do I have to do to achieve this:

$ pwd
/tmp
$ python baz.py
running from /tmp 
file is baz.py

__file__ is NOT what you are looking for. Don't use accidental side-effects

sys.argv[0] is always the path to the script (if in fact a script has been invoked) -- see http://docs.python.org/library/sys.html#sys.argv

__file__ is the path of the currently executing file (script or module). This is accidentally the same as the script if it is accessed from the script! If you want to put useful things like locating resource files relative to the script location into a library, then you must use sys.argv[0].

Example:

C:\junk\so>type \junk\so\scriptpath\script1.py
import sys, os
print "script: sys.argv[0] is", repr(sys.argv[0])
print "script: __file__ is", repr(__file__)
print "script: cwd is", repr(os.getcwd())
import whereutils
whereutils.show_where()

C:\junk\so>type \python26\lib\site-packages\whereutils.py
import sys, os
def show_where():
    print "show_where: sys.argv[0] is", repr(sys.argv[0])
    print "show_where: __file__ is", repr(__file__)
    print "show_where: cwd is", repr(os.getcwd())

C:\junk\so>\python26\python scriptpath\script1.py
script: sys.argv[0] is 'scriptpath\\script1.py'
script: __file__ is 'scriptpath\\script1.py'
script: cwd is 'C:\\junk\\so'
show_where: sys.argv[0] is 'scriptpath\\script1.py'
show_where: __file__ is 'C:\\python26\\lib\\site-packages\\whereutils.pyc'
show_where: cwd is 'C:\\junk\\so'