在 python setup.py data_files 中包含整个目录 [英] Include entire directory in python setup.py data_files

问题描述

设置的 data_files 参数采用以下格式的输入:

The data_files parameter for setup takes input in the following format:

setup(...
    data_files = [(target_directory, [list of files to be put there])]
    ....)

有没有办法让我指定整个数据目录，这样我就不必单独命名每个文件并在更改项目中的实现时更新它?

Is there a way for me to specify an entire directory of data instead, so I don't have to name each file individually and update it as I change implementation in my project?

我尝试使用 os.listdir()，但我不知道如何使用相对路径来做到这一点，我无法使用 os.getcwd() 或 os.realpath(__file__) 因为它们没有正确指向我的存储库根目录.

I attempted to use os.listdir(), but I don't know how to do that with relative paths, I couldn't use os.getcwd() or os.realpath(__file__) since those don't point to my repository root correctly.

推荐答案

我不知道如何使用相对路径来做到这一点

I don't know how to do that with relative paths

需要先获取目录的路径，所以...

You need to get the path of the directory first, so...

假设你有这个目录结构:

Say you have this directory structure:

cur_directory
|- setup.py
|- inner_dir
   |- file2.py

要获取当前文件的目录(在本例中为 setup.py)，请使用:

To get the directory of the current file (in this case setup.py), use this:

cur_directory_path = os.path.abspath(os.path.dirname(__file__))

然后，要获得相对于current_directory的目录路径，只需加入一些其他目录，例如:

Then, to get a directory path relative to current_directory, just join some other directories, eg:

inner_dir_path = os.path.join(cur_directory_path, 'inner_dir')

如果你想上移一个目录，只需使用.."，例如:

If you want to move up a directory, just use "..", for example:

parent_dir_path = os.path.join(current_directory, '..')

一旦你有了那个路径，你就可以做 os.listdir

Once you have that path, you can do os.listdir

为了完整性:

如果你想要一个文件的路径，在这种情况下file2.py"相对于setup.py，你可以这样做:

If you want the path of a file, in this case "file2.py" relative to setup.py, you could do:

file2_path = os.path.join(cur_directory_path, 'inner_dir', 'file2.py') 

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