假设工程根目录为 d:/Project/
然后在 d:/Project/a/b/c/d/file.py 里使用 os.getcwd() 方法获取的是
d:/Project/a/b/c/d
现在我想获取 d:/Project 怎么做?
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谢谢大家的热情回答,都怪我没有将问题说清楚,我的意思是获取当前工程的根目录,而获取根目录的函数可能在任何目录下
根目录既可能是 d:/Project/
也可能是 d:/Python/Project/
也可能是 d:/balabala/Python/Project/
因此逐级往上也是不可能找到的,我目前的方法是在根目录下放一个文件 file
然后这样:
def getSeparator():
if 'Windows' in platform.system():
separator = '\\'
else:
separator = '/'
return separator
def findPath(file):
o_path = os.getcwd()
separator = getSeparator()
str = o_path
str = str.split(separator)
while len(str) > 0:
spath = separator.join(str)+separator+file
leng = len(str)
if os.path.exists(spath):
return spath
str.remove(str[leng-1])
但是这样也有一个问题,就是必须保证任何目录下不能有和根目录下file重名的文件
黄舟2017-04-17 17:48:09
There is no good way unless your Project directory is never moved.
Add a module in the Project directory to get the path of the current fileproject_dir = os.path.dirname(os.path.abspath(__file__))
, and then import it in file.py
高洛峰2017-04-17 17:48:09
Go up step by step
os.path.dirname(os.path.dirname(os.path.dirname(os.path.dirname(os.getcwd()))))
阿神2017-04-17 17:48:09
If you finally mobilize file.py's os.getcwd()
under the Project file, you will get the result of D:/Project
. os.getcwd()
的话,就会得到D:/Project
这个结果。
比如说你建立了一个这样的项目结构,如果在是在D:/Project/a/b/c/d/file.py
直接调用的话,很自然就是得file.py
文件所在的工作目录的路径。
如果是在D:/Project/__init__.py
调用呢,那就是得到D:/Project/__init__.py
所在的工作目录的路径,也就是D:/Project/
了。
os.getcwd()
D:/Project/a/b/c/d/file.py
, it will naturally be file.py
The path to the working directory where the file is located. If it is called from D:/Project/__init__.py
, then the path to the working directory where D:/Project/__init__.py
is located is obtained, that is, < code>D:/Project/.
os.getcwd()
The results obtained are different depending on the path of the calling file. 🎜
🎜
🎜If you are sure of the name of the project directory and ensure that no folder with the same name will appear in the project directory, you can use regular expressions. 🎜
import re, os
# 测试
paths = ['d:\Project\', 'home/Python/Project/', 'c:/balabala/Python/Project/']
for path in paths:
pj_dir = re.match('.*Project', path)
print(pj_dir.group())
# 在子文件下就应该这样用
print(re.match('(.*\{sep}Project)\{sep}'.format(sep=os.sep), __file__).group(1))
高洛峰2017-04-17 17:48:09
import os
s1 = "d:/Project/a/b/c/d/file.py"
s2 = r"d:\project\a\b\c\d\file.py"
for i in [s1, s2]:
abs_path = os.path.abspath(i).split(os.sep)
print os.path.abspath(abs_path[0] + os.sep + abs_path[1])
Directory separator for "" and "/"