将xlsx上载到Google云端硬盘共享文件夹

Question

I would like to upload the file risk_pos.xlsx which is in the same directory as the .py file, into a shared drive. Any help would be appreciated please.

My code:

scope = ["https://spreadsheets.google.com/feeds",'https://www.googleapis.com/auth/spreadsheets',"https://www.googleapis.com/auth/drive.file","https://www.googleapis.com/auth/drive"]
os.chdir(sys.path[0])
creds = ServiceAccountCredentials.from_json_keyfile_name("client_secret1.json", scope)
client = gspread.authorize(creds)

DRIVE = build('drive', 'v3', http=creds.authorize(Http()))

files = DRIVE.files().list().execute().get('files', [])
for f in files:
    print(f['name'], f['mimeType'])


file_metadata = {'inventory': 'risk_pos.xlsx', 'parents': ['XXXXXXXXXXXXXXXX']}
media = MediaFileUpload('risk_pos.xlsx', mimetype='application/vnd.ms-excel')
files = DRIVE.files().create(body=file_metadata, supportsTeamDrives=True, media_body=media, fields='id').execute()

The error:

File "risk_pos3.py", line 59, in <module>
    files = DRIVE.files().create(body=file_metadata, supportsTeamDrives=True, media_body=media, fields='id').execute()
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/googleapiclient/_helpers.py", line 130, in positional_wrapper
return wrapped(*args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/site-packages/googleapiclient/http.py", line 851, in execute
raise HttpError(resp, content, uri=self.uri)
googleapiclient.errors.HttpError: <HttpError 404 when requesting 
https://www.googleapis.com/upload/drive/v3/files?supportsTeamDrives=true&fields=id&alt=json&uploadType=multipart returned "File not found: XXXXXXXXXXXXXXXXXX.">

EDIT:

I don't have error now. I just removed the brackets around the 'parents': ['XXXXXXXXXXXXXXXX'] But I don't see the file showing up in Drive. Still every time i run the process it seems that something new appear. A file named untitled... as below you can see the number of file increasing. But no untitled... files in Drive.

 JRACMB-MahfoudD:market_risk delalma$ python3 risk_pos3.py 
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Market_data application/vnd.google-apps.spreadsheet
 Market_data_volume application/vnd.google-apps.spreadsheet
 Getting started application/pdf
JRACMB-MahfoudD:market_risk delalma$ python3 risk_pos3.py 
Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Market_data application/vnd.google-apps.spreadsheet
 Market_data_volume application/vnd.google-apps.spreadsheet
 Getting started application/pdf
 JRACMB-MahfoudD:market_risk delalma$ python3 risk_pos3.py 
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Untitled application/vnd.ms-excel
 Market_data application/vnd.google-apps.spreadsheet
 Market_data_volume application/vnd.google-apps.spreadsheet
 Getting started application/pdf

Answer 1

Service accounts are tricky. If there is no error with the upload then its probably being uploaded the question is where is it being uploaded to?

A service account is not you its a dummy user with its own drive account. You may have uploaded the file to the service accounts drive account. The first thing i would do is run a file.list can you see the file then?

The thing is you say that you are using parents in your request. Which means either you created a new folder on the service account and are using that, or you have shared a folder on your personal drive account and are trying to upload to that. Either way a file.list will help you figure out where the file is if the service account uploaded it it has access to it its going to be the best easiest way to find the file.

untitled

Files are uploaded in two parts. First the metadata containing things like the files name and second the actual data of the file itself. Your problem is that you are not uploading the metadata.

file_metadata = {'name': 'photo.jpg'}
media = MediaFileUpload('files/photo.jpg',
                        mimetype='image/jpeg')
file = drive_service.files().create(body=file_metadata,
                                    media_body=media,
                                    fields='id').execute()
print 'File ID: %s' % file.get('id')

You may want to read managing uploads

As for the always creating a new file that is logical considering that you are sending file.create then a new file will always be created if you want it to over write an existing file then you will need to do file.update