[英]full file path for zipped folder in python
我有以下代码试图获取完整的文件路径(包括文件夹):
import zipfile
import os
import sys
zipped_files_dir = 'Z:\Dev\some_files'
def get_folder_names():
path_list = []
for folder_name in os.listdir(zipped_files_dir):
path_list.append(folder_name)
return path_list
def get_folder_directories(folder_list):
for folder in folder_list:
pathname = os.path.abspath(folder)
print(pathname)
def main():
get_folder_directories(get_folder_names())
>>>Z:\Dev\new_folder.zip
我的问题是我应该在返回的目录中有“\\ some_files \\ new_folder.zip”。 有任何想法吗?
谢谢!
您可以在get_folder_directories
函数中使用os.path.join(zipped_files_dir, folder)
:
import zipfile
import os
import sys
zipped_files_dir = 'Z:\Dev\some_files'
def get_folder_names():
path_list = []
for folder_name in os.listdir(zipped_files_dir):
path_list.append(folder_name)
return path_list
def get_folder_directories(folder_list):
for folder in folder_list:
pathname = os.path.abspath(os.path.join(zipped_files_dir, folder))
print(pathname)
def main():
get_folder_directories(get_folder_names())
或者,在get_folder_names()
中get_folder_names()
path_list.append(os.path.join(zipped_files_dir, folder_name))
get_folder_names()
:
import zipfile
import os
import sys
zipped_files_dir = 'Z:\Dev\some_files'
def get_folder_names():
path_list = []
for folder_name in os.listdir(zipped_files_dir):
path_list.append(os.path.join(zipped_files_dir, folder_name))
return path_list
def get_folder_directories(folder_list):
for folder in folder_list:
pathname = os.path.abspath(folder)
print(pathname)
def main():
get_folder_directories(get_folder_names())
Chown有正确的解决方案。
在你的代码中,你传递abs_path字符串“new_folder.zip”。 但是abs_path不知道它来自哪里,因此它确定它必须在当前工作目录中,这就是为什么您得到r“ Z:\\ Dev \\ new_folder.zip”的原因。 您需要使用os.path.join将文件名与您在其中找到的路径组合在一起。
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