如何使用 sys.path.append 在 python 中导入文件？
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How to import files in python using sys.path.append?
提问by Hunle
There are two directories on my desktop, DIR1and DIR2which contain the following files:
有我的桌面上两个目录，DIR1并且DIR2包含下列文件：
DIR1:
file1.py
DIR2:
file2.py myfile.txt
The files contain the following:
这些文件包含以下内容：
file1.py
文件1.py
import sys
sys.path.append('.')
sys.path.append('../DIR2')
import file2
file2.py
文件2.py
import sys
sys.path.append( '.' )
sys.path.append( '../DIR2' )
MY_FILE = "myfile.txt"
myfile = open(MY_FILE)
myfile.txt
我的文件.txt
some text
Now, there are two scenarios. The first works, the second gives an error.
现在，有两种情况。第一个有效，第二个给出错误。
Scenario 1
场景一
I cdinto DIR2and run file2.pyand it runs no problem.
我cd进入DIR2并运行file2.py，它运行没有问题。
Scenario 2
场景二
I cdinto DIR1and run file1.pyand it throws an error:
我cd进入DIR1并运行file1.py它会引发错误：
Traceback (most recent call last):
File "<absolute-path>/DIR1/file1.py", line 6, in <module>
import file2
File "../DIR2/file2.py", line 9, in <module>
myfile = open(MY_FILE)
IOError: [Errno 2] No such file or directory: 'myfile.txt'
However, this makes no sense to me, since I have appended the path to file1.pyusing the command sys.path.append('../DIR2').
但是，这对我来说没有意义，因为我已经附加了file1.py使用命令的路径sys.path.append('../DIR2')。
Why does this happen when file1.py, when file2.pyis in the same directory as myfile.txtyet it throws an error? Thank you.
为什么会在file1.py, whenfile2.py位于同一目录中时发生这种情况，myfile.txt但它会引发错误？谢谢你。
采纳答案by larsks
You can create a path relative to a module by using a module's __file__attribute. For example:
您可以使用模块的__file__属性创建相对于模块的路径。例如：
myfile = open(os.path.join(
os.path.dirname(__file__),
MY_FILE))
This should do what you want regardless of where you start your script.
无论您从哪里开始脚本，这都应该做您想做的事情。
回答by demented hedgehog
Replace
代替
MY_FILE = "myfile.txt"
myfile = open(MY_FILE)
with
和
MY_FILE = os.path.join("DIR2", "myfile.txt")
myfile = open(MY_FILE)
That's what the comments your question has are referring to as the relative path solution. This assumes that you're running it from the dir one up from myfile.txt... so not ideal.
这就是您的问题所指的相对路径解决方案的评论。这假设您从 myfile.txt 的目录中运行它......所以不理想。
If you know that my_file.txt is always going to be in the same dir as file2.py then you can try something like this in file2..
如果您知道 my_file.txt 始终与 file2.py 位于同一目录中，那么您可以在 file2..
from os import path
fname = path.abspath(path.join(path.dirname(__file__), "my_file.txt"))
myfile = open(fname)
