到目前为止,我可以将文件上传到该文件夹(如果存在)。但我无法找到一种方法来创建它。因此,如果该文件夹不存在,我的脚本就会终止。
import sys
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
gpath = '2015'
fname = 'Open Drive Replacements 06_01_2015.xls'
gauth = GoogleAuth()
gauth.LocalWebserverAuth()
drive = GoogleDrive(gauth)
file_list = drive.ListFile({'q': "'root' in parents and trashed=false"}).GetList()
for file1 in file_list:
if file1['title'] == gpath:
id = file1['id']
file1 = drive.CreateFile({'title': fname, "parents": [{"kind": "drive#fileLink","id": id}]})
file1.SetContentFile(fname)
file1.Upload()
如果gpath不存在,你能帮我修改上面的代码来创建文件夹吗?
基于文档 https://developers.google.com/drive/web/folder, 它应该是
file1 = drive.CreateFile({'title': fname,
"parents": [{"id": id}],
"mimeType": "application/vnd.google-apps.folder"})
Update:截至 2020 年 4 月,文档 https://developers.google.com/drive/api/v3/folder#python_1(v3) 已更新 API 文档并显示:
folder_id = '0BwwA4oUTeiV1TGRPeTVjaWRDY1E'
file_metadata = {
'name': 'photo.jpg',
'parents': [folder_id]
}
media = MediaFileUpload('files/photo.jpg',
mimetype='image/jpeg',
resumable=True)
file = drive_service.files().create(body=file_metadata,
media_body=media,
fields='id').execute()
print 'File ID: %s' % file.get('id')
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