我从PyDrive文档中获得以下代码，该代码允许访问我的Google云端硬盘中的顶级文件夹。我想从中访问所有文件夹，子文件夹和文件。我该怎么做呢（我刚刚开始使用PyDrive）？
#!/usr/bin/python
# -*- coding: utf-8 -*-
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
gauth = GoogleAuth()
gauth.LocalWebserverAuth() # Creates local webserver and auto handles authentication
#Make GoogleDrive instance with Authenticated GoogleAuth instance
drive = GoogleDrive(gauth)
#Google_Drive_Tree =
# Auto-iterate through all files that matches this query
top_list = drive.ListFile({'q': "'root' in parents and trashed=false"}).GetList()
for file in top_list:
print 'title: %s, id: %s' % (file['title'], file['id'])
print "---------------------------------------------"
#Paginate file lists by specifying number of max results
for file_list in drive.ListFile({'q': 'trashed=true', 'maxResults': 10}):
print 'Received %s files from Files.list()' % len(file_list) # <= 10
for file1 in file_list:
print 'title: %s, id: %s' % (file1['title'], file1['id'])
我已经查看了以下页面How to list all files, folders, subfolders and subfiles of a Google drive folder，这似乎是我正在寻找的答案，但代码已经不存在了。
答案 0 :(得分：4)
它需要迭代文件列表。基于this，代码获取文件夹中文件的标题和每个文件的URL链接。代码可调，以通过提供@Override protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (resultCode != Activity.RESULT_OK) return;
if (requestCode == AppConstants.REQUEST_CAMERA_PICK) {
Bitmap photo = (Bitmap) data.getExtras().get("data");
ByteArrayOutputStream stream = new ByteArrayOutputStream();
photo.compress(Bitmap.CompressFormat.PNG, 100, stream);
Glide.with(this)
.load(stream.toByteArray())
.asBitmap()
.error(R.drawable.ic_thumb_placeholder)
.transform(new CircleTransform(this))
.into(imageview);
}
}
等文件夹的id来获取特定文件夹。下面给出的示例是查询ListFolder('id')
root答案 1 :(得分：4)
您的代码绝对正确。但是使用Pydrive的默认设置，您只能访问根级文件和文件夹。 在settings.yaml文件中更改oauth_scope可修复此问题。
client_config_backend: settings
client_config:
client_id: XXX
client_secret: XXXX
save_credentials: True
save_credentials_backend: file
save_credentials_file: credentials.json
get_refresh_token: True
oauth_scope:
- https://www.googleapis.com/auth/drive
- https://www.googleapis.com/auth/drive.metadata
答案 2 :(得分：1)
这是我对获取子文件夹中所有文件的看法... 这使您可以通过设置的路径进行查询。这是不同的，因为它不会对每个文件夹发出 1 个请求。它会批量创建要查询的文件夹。
批处理代码段：
'some_id_1234' in parents or 'some_id_1235' in parents or 'some_id_1236' in parents or 'some_id_1237' in parents or 'some_id_1238' in parents or 'some_id_1239' in parents or 'some_id_1240' in parents and trashed=false
然后您可以一次查询多个文件夹中的文件。您的查询大小不能太大，因此任何超过 300 多个文件夹（'some_id_1234' in parents'），您就会开始出错，因此将批量大小保持在 250 左右。 >
假设您要检查的文件夹有 1,110 个文件夹，并且您将批量大小设置为 250。 然后它将发出 5 个单独的请求来查询所有文件夹。
-Request 1 查询 250 个文件夹
-Request 2 查询 250 个文件夹
-Request 3 查询 250 个文件夹
-Request 4 查询 250 个文件夹
-Request 5 查询 110 个文件夹
然后里面的任何子文件夹都会批量创建并递归查询。
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
def parse_gdrive_path(gd_path):
if ':' in gd_path:
gd_path = gd_path.split(':')[1]
gd_path = gd_path.replace('\\', '/').replace('//', '/')
if gd_path.startswith('/'):
gd_path = gd_path[1:]
if gd_path.endswith('/'):
gd_path = gd_path[:-1]
return gd_path.split('/')
def resolve_path_to_id(folder_path):
_id = 'root'
folder_path = parse_gdrive_path(folder_path)
for idx, folder in enumerate(folder_path):
folder_list = gdrive.ListFile({'q': f"'{_id}' in parents and title='{folder}' and trashed=false and mimeType='application/vnd.google-apps.folder'", 'fields': 'items(id, title, mimeType)'}).GetList()
_id = folder_list[0]['id']
title = folder_list[0]['title']
if idx == (len(folder_path) - 1) and folder == title:
return _id
return _id
def get_folder_files(folder_ids, batch_size=100):
base_query = "'{target_id}' in parents"
target_queries = []
query = ''
for idx, folder_id in enumerate(folder_ids):
query += base_query.format(target_id=folder_id)
if len(folder_ids) == 1 or idx > 0 and idx % batch_size == 0:
target_queries.append(query)
query = ''
elif idx != len(folder_ids)-1:
query += " or "
else:
target_queries.append(query)
for query in target_queries:
for f in gdrive.ListFile({'q': f"{query} and trashed=false", 'fields': 'items(id, title, mimeType, version)'}).GetList():
yield f
def get_files(folder_path=None, target_ids=None, files=[]):
if target_ids is None:
target_ids = [resolve_path_to_id(folder_path)]
file_list = get_folder_files(folder_ids=target_ids, batch_size=250)
subfolder_ids = []
for f in file_list:
if f['mimeType'] == 'application/vnd.google-apps.folder':
subfolder_ids.append(f['id'])
else:
files.append(f['title'])
if len(subfolder_ids) > 0:
get_files(target_ids=subfolder_ids)
return files
gauth = GoogleAuth()
gauth.LocalWebserverAuth()
gdrive = GoogleDrive(gauth)
file_list = get_files('/Some/Folder/Path')
for f in file_list:
print(f)
例如：
您的谷歌驱动器包含以下内容：
(folder) Root
(folder) Docs
(subfolder) Notes
(subfolder) School
(file) notes_1.txt
(file) notes_2.txt
(file) notes_3.txt
(file) notes_4.txt
(file) notes_5.txt
(subfolder) Important
(file) important_notes_1.txt
(file) important_notes_2.txt
(file) important_notes_3.txt
(subfolder) Old Notes
(file) old_1.txt
(file) old_2.txt
(file) old_3.txt
(subfolder) Secrets
(file) secret_1.txt
(file) secret_2.txt
(file) secret_3.txt
(folder) Stuff
(file) nothing.txt
(file) this-will-not-be-found.txt
并且您想从“Notes”文件夹/子文件夹中获取所有文件
你会这样做：
file_list = get_files('/Docs/Notes')
for f in file_list:
print(f)
Output:
>> notes_1.txt
>> notes_2.txt
>> notes_3.txt
>> notes_4.txt
>> notes_5.txt
>> important_notes_1.txt
>> important_notes_2.txt
>> important_notes_3.txt
>> old_1.txt
>> old_2.txt
>> old_3.txt
>> secret_1.txt
>> secret_2.txt
>> secret_3.txt
希望这对某人有所帮助:)