Python:如果目录是符号链接，则getcwd和pwd给出不同的结果 [英] Python : getcwd and pwd if directory is a symbolic link give different results

问题描述

如果我的工作目录是符号链接，则os.getcwd()和os.system("pwd")不会给出相同的结果.我想使用os.path.abspath(".")故意获取我的工作目录(或其中的文件)的完整路径，而不是获得与os.path.realpath(".")相同的结果.

If my working directory is a symbolic link, os.getcwd() and os.system("pwd") do not give the same result. I would like to use os.path.abspath(".") to get the full path of my working directory (or a file in it), on purpose, not to get the same result than os.path.realpath(".").

如何在python 2.7中获取类似os.path.abspath(".", followlink=False)的东西?

How to get, in python 2.7, something like os.path.abspath(".", followlink=False) ?

示例:/tmp是/private/tmp的符号链接

Example : /tmp is a symbolic link to /private/tmp

cd /tmp
touch toto.txt
python
print os.path.abspath("toto.txt")
--> "/private/tmp/toto.txt"
os.system("pwd")
--> "/tmp"
os.getcwd()
--> "/private/tmp"

如何从相对路径"toto.txt"中获取"/tmp/toto.txt"?

How can I get "/tmp/toto.txt" from the relative path "toto.txt" ?

推荐答案

解决方案是:

from subprocess import Popen, PIPE

def abspath(pathname):
    """ Returns absolute path not following symbolic links. """
    if pathname[0]=="/":
        return pathname
    # current working directory
    cwd = Popen("pwd", stdout=PIPE, shell=True).communicate()[0].strip()
    return os.path.join(cwd, pathname)

print os.path.abspath("toto.txt")  # --> "/private/tmp/toto.txt"
print abspath("toto.txt")          # --> "/tmp/toto.txt"

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