我想采用任意数量的路径来表示嵌套的tar档案,并对最里面的档案执行操作.问题是,嵌套可以是任意的,因此我需要的上下文管理器的数量也是任意的.
举个例子：
ARCHIVE_PATH = "path/to/archive.tar"INNER_PATHS = ("nested/within/archive/one.tar","nested/within/archive/two.tar",# Arbitary number of these)def list_inner_contents(archive_path,inner_paths):with TarFile(archive_path) as tf1:with TarFile(fileobj=tf1.extractfile(inner_paths[0])) as tf2:with TarFile(fileobj=tf2.extractfile(inner_paths[1])) as tf3:# ...arbitary level of these!return tfX.getnames()contents = list_inner_contents(ARCHIVE_PATH,INNER_PATHS))
我不能使用with语句的nesting syntax,因为可以有任意数量的级别来嵌套.我不能使用contextlib.nested因为文档在那里说：
…using
nested()to open two files is a programming error as the first file will not be closed promptly if an exception is thrown when opening the second file.
有没有办法使用语言结构来执行此操作,还是需要手动管理我自己的打开文件对象堆栈？
最佳答案
对于这种情况,您可以使用递归.对于这种情况来说感觉最自然(当然如果在Python中没有特殊处理)：
ARCHIVE_PATH = "path/to/archive.tar"INNER_PATHS = ["nested/within/archive/one.tar",# Arbitary number of these]def list_inner_contents(archive_path,inner_paths):def rec(tf,rest_paths):if not rest_paths:return tf.getnames()with TarFile(fileobj=tf.extractfile(rest_paths[0])) as tf2:return rec(tf2,rest_paths[1:])with TarFile(archive_path) as tf:try:return rec(tf,inner_paths)except RuntimeError:# We come here in case the inner_paths list is too long# and we go too deeply in the recursionreturn None