如何在Python中递归解决目录路径问题？

Question

我正在执行 REST API 调用以获取 SharePoint 文档库的文件夹。

我想递归获取整个目录树中的所有文件夹路径。

我编写了一个函数来从给定文件夹中获取子文件夹列表，但我不确定如何遍历到第 N 个目录并获取所有文件夹路径。

例如，假设当前的 SharePoint 文档库结构如下 JSON (fo=folder; f=file)：

{
  "root": [
    {
      "fo1": {
        "fo1": "f1",
        "fo2": ["f1", "f2"]
      },
      "fo2": ["fi1", "fi2"]
    },
    "fi1","fi2"]
}

从上面的例子中，我想要一个所有文件夹/目录的路径列表： 例如输出应该是：

["/root/fo1/", "/root/fo1/fo1/", "/root/fo1/fo2/", "/root/fo2/"]

因为它是一个 REST API 调用，所以我事先不知道结构，直到我运行获取子文件夹的查询，然后进入每个子文件夹以获取它们各自的子文件夹。

我编写的当前（以下）函数正在获取数据直到 1 级（子文件夹，因为它是基于内部迭代而不是递归的），我如何实现基于递归的解决方案来获取所有唯一的文件夹路径为列表？

def print_root_contents(ctx):

    try:
        list_object = ctx.web.lists.get_by_title('Documents')
        folder = list_object.root_folder
        ctx.load(folder)
        ctx.execute_query()

        folders = folder.folders
        ctx.load(folders)
        ctx.execute_query()

        for myfolder in folders:
            print("For Folder : {0}".format(myfolder.properties["Name"]))
            folder_list, files_list = print_folder_contents(ctx, myfolder.properties["Name"])
            print("Sub folders - ", folder_list)
            print("Files - ", files_list)

    except Exception as e:
        print('Problem printing out library contents: ', e)


def print_folder_contents(ctx, folder_name):

    try:
        folder = ctx.web.get_folder_by_server_relative_url("/sites/abc/Shared Documents/"+folder_name+"/")
        ctx.load(folder)
        ctx.execute_query()

        # Folders
        fold_names = []
        sub_folders = folder.folders
        ctx.load(sub_folders)
        ctx.execute_query()
        for s_folder in sub_folders:
            # folder_name = folder_name+"/"+s_folder.properties["Name"]
            # print("Folder name: {0}".format(folder.properties["Name"]))
            fold_names.append(s_folder.properties["Name"])

        return fold_names

    except Exception as e:
        print('Problem printing out library contents: ', e)

在上面的最后一个函数 (print_folder_contents) 中，我无法形成递归逻辑来保持递归地附加文件夹和子文件夹，并在第 n 个文件夹中没有更多文件夹时停止它并继续下一个同级文件夹一级起来。

发现它真的很有挑战性。有什么帮助吗？

Answer 1

您可以使用生成器函数迭代 dict 项并生成 dict 键和 yield 键，并与递归调用生成的路径连接，如果给定一个列表，则递归地生成从列表项上的递归调用生成的内容：

def paths(d):
    def _paths(d):
        if isinstance(d, dict):
            for k, v in d.items():
                yield k + '/'
                for p in _paths(v):
                    yield '/'.join((k, p))
        elif isinstance(d, list):
            for i in d:
                yield from _paths(i)
    return ['/' + p for p in _paths(d)]

所以给定：

d = {
  "root": [
    {
      "fo1": {
        "fo1": "f1",
        "fo2": ["f1", "f2"]
      },
      "fo2": ["fi1", "fi2"]
    },
    "fi1","fi2"]
}

paths(d) 返回：

['/root/', '/root/fo1/', '/root/fo1/fo1/', '/root/fo1/fo2/', '/root/fo2/']

请注意，您的预期输出应包括 '/root/'，因为根文件夹也应该是有效文件夹。

Answer 2

我知道这个答案对游戏来说已经很晚了，但是您可以执行以下操作来获取给定某个父目录的所有子 SharePoint 对象的平面列表。

这是因为我们不断扩展单个列表，而不是在递归某些目录树时利用 list.append() 方法创建嵌套对象。

我相信会有机会改进以下 sn-p，但我相信这应该有助于您实现目标。

干杯，

rs311

from office365.sharepoint.client_context import ClientContext


def get_items_in_directory(ctx_client: ClientContext,
                           directory_relative_uri: str,
                           recursive: bool = True):
    """
    This function provides a way to get all items in a directory in SharePoint, with
    the option to traverse nested directories to extract all child objects.
    
    :param ctx_client: office365.sharepoint.client_context.ClientContext object
        SharePoint ClientContext object.
    :param directory_relative_uri: str
        Path to directory in SharePoint. 
    :param recursive: bool
        default = False
        Tells function whether or not to perform a recursive call.
    :return: list
        Returns a flattened array of all child file and/or folder objects
        given some parent directory. All items will be of the following types:
            - office365.sharepoint.file.File
            - office365.sharepoint.folder.Folder
        
    Examples 
    ---------
    All examples assume you've already authenticated with SharePoint per
    documentation found here:
        - https://github.com/vgrem/Office365-REST-Python-Client#examples
        
    Assumed directory structure:
        some_directory/
            my_file.csv
            your_file.xlsx
            sub_directory_one/
                123.docx
                abc.csv
            sub_directory_two/
                xyz.xlsx
    
    directory = 'some_directory'
    # Non-recursive call
    extracted_child_objects = get_items_in_directory(directory)
    # extracted_child_objects would contain (my_file.csv, your_file.xlsx, sub_directory_one/, sub_directory_two/)
    
    
    # Recursive call
    extracted_child_objects = get_items_in_directory(directory, recursive=True)
    # extracted_child_objects would contain (my_file.csv, your_file.xlsx, sub_directory_one/, sub_directory_two/, sub_directory_one/123.docx, sub_directory_one/abc.csv, sub_directory_two/xyz.xlsx)
    
    """
    contents = list()
    folders = ctx_client.web.get_folder_by_server_relative_url(directory_relative_uri).folders
    ctx_client.load(folders)
    ctx_client.execute_query()

    if recursive:
        for folder in folders:
            contents.extend(
                get_items_in_directory(
                    ctx_client=ctx_client,
                    directory_relative_uri=folder.properties['ServerRelativeUrl'],
                    recursive=recursive)
            )

    contents.extend(folders)

    files = ctx_client.web.get_folder_by_server_relative_url(directory_relative_uri).files
    ctx_client.load(files)
    ctx_client.execute_query()

    contents.extend(files)
    return contents