I have a GCP project and there is a Jupiter notebook which is to read files from G Drive.
def search_file(request = None):
try:
service = build('drive', 'v3', credentials=creds)
# Call the Drive v3 API
results = service.files().list(
pageSize=1, fields="nextPageToken, files(id, name)").execute()
items = results.get('files', [])
if not items:
print('No files found.')
return
print('Files:')
for item in items:
print(u'{0} ({1})'.format(item['name'], item['id']))
except HttpError as error:
print(f'An error occurred: {error}')
I'm using Google's quickstart to retrieve files from G Drive. The authentication and all works fine and the intended drive has just one file on it. But when I execute this search_file method, I'm getting No Files found. message. I'm trying to read one csv file which resides in G Drive. What I have missed or any other alternatives for this?
Can you try running the query in the following format?
shared_drive_id = "" # you need to specify specific id of drive folder provided in url of that folder such as https://drive.google.com/drive/folders/<shared_drive_id>
query_file = "mimeType='application/vnd.google-apps.document' and trashed=false"
search_file = drive_service.files().list(q=query_file,
fields='files(id)',
driveId=shared_drive_id,
supportsAllDrives=True,
includeItemsFromAllDrives=True,
corpora='drive').execute()
files = search_file.get('files', [])
Related
how are you doing?
I'm trying to rename some files in my Google Drive folder, following the instructions listed in documentation.
However I'm getting an error in one of the arguments (FileNotFoundError: [Errno 2] No such file or directory: 'trying_new_filename_string')
I ONLY NEED TO RENAME THE DOCUMENT, NOTHING ELSE.
This is my code:
service = build('drive', 'v3', credentials=credentials)
file_id = '1TKxcYDzEK3SUjSv6dCMM6WkKRmZwm84SPWVhR1F2DEc'
new_title = 'trying_new_title'
new_mime_type = 'trying_new_mime_type'
new_filename = 'trying_new_filename_string'
from apiclient import errors
from apiclient.http import MediaFileUpload
def update_file(service, file_id, new_title, new_mime_type, new_filename):
# First retrieve the file from the API.
file = service.files().get(fileId=file_id).execute()
# File's new metadata.
file['name'] = new_title
file['mimeType'] = new_mime_type
# File's new content.
media_body = MediaFileUpload(
new_filename, mimetype=new_mime_type, resumable=True)
# Send the request to the API.
updated_file = service.files().update(
fileId=file_id,
body=file,
media_body=media_body).execute()
return updated_file
Seems related to the fact the new_filename does not exist. As the MediaUploadFile tries to open and does not find a file with the trying_new_filename_string name.
As you only need to rename that file, don't change its content, you should remove the MediaUploadFile method.
Try:
def renameFile(service, fileId, newTitle):
body = {'name': newTitle}
return service.files().update(fileId=fileId, body=body).execute()
I exported some images from Google Earth Engine to Google Drive. I need to download those images to a local drive using a Python script. Then, I tried to use oauth2client, apiclient as I saw here:
I got a list of files in Drive and the corresponding IDs, then I use the ID to try to download the file using the gdown lib:
gdown.download(f'https://drive.google.com/uc?id={file_data["id"]}',
f'{download_path}{os.sep}{filename_to_download}.tif')
I got the following error message:
Access denied with the following error:
Cannot retrieve the public link of the file. You may need to change
the permission to 'Anyone with the link', or have had many accesses.
You may still be able to access the file from the browser:
https://drive.google.com/uc?id=<id>
As I got the Drive file list, I suppose that the Drive authentication is ok. If I use the error message suggested link in the browser, I can download the file. If a check file properties at Drive, I can see:
Who can access: not shared.
What should I do to download the files?
This is the complete code:
# https://medium.com/swlh/google-drive-api-with-python-part-i-set-up-credentials-1f729cb0372b
# https://levelup.gitconnected.com/google-drive-api-with-python-part-ii-connect-to-google-drive-and-search-for-file-7138422e0563
# https://stackoverflow.com/questions/38511444/python-download-files-from-google-drive-using-url
import os
from apiclient import discovery
from httplib2 import Http
from oauth2client import client, file, tools
import gdown
class GoogleDrive(object):
# define API scope
def __init__(self, secret_credentials_file_path = './credentials'):
self.DriveFiles = None
SCOPE = 'https://www.googleapis.com/auth/drive'
self.store = file.Storage(f'{secret_credentials_file_path}{os.sep}credentials.json')
self.credentials = self.store.get()
if not self.credentials or self.credentials.invalid:
flow = client.flow_from_clientsecrets(f'{secret_credentials_file_path}{os.sep}client_secret.json',
SCOPE)
self.credentials = tools.run_flow(flow, self.store)
oauth_http = self.credentials.authorize(Http())
self.drive = discovery.build('drive', 'v3', http=oauth_http)
def RetrieveAllFiles(self):
results = []
page_token = None
while True:
try:
param = {}
if page_token:
param['pageToken'] = page_token
files = self.drive.files().list(**param).execute()
# append the files from the current result page to our list
results.extend(files.get('files'))
# Google Drive API shows our files in multiple pages when the number of files exceed 100
page_token = files.get('nextPageToken')
if not page_token:
break
except Exception as error:
print(f'An error has occurred: {error}')
break
self.DriveFiles = results
def GetFileData(self, filename_to_search):
for file_data in self.DriveFiles:
if file_data.get('name') == filename_to_search:
return file_data
else:
return None
def DownloadFile(self, filename_to_download, download_path):
file_data = self.GetFileData(f'{filename_to_download}.tif')
gdown.download(f'https://drive.google.com/uc?id={file_data["id"]}',
f'{download_path}{os.sep}{filename_to_download}.tif')
Google drive may not be the best tool for this, you may want to upload them into a RAW file hosting service like Imgur and download it to a file using requests, you can then read the file using the script or you don't even have to write it to the file and just use image.content instead to specify the image. Here's an example:
image = requests.get("https://i.imgur.com/5SMNGtv.png")
with open("image.png", 'wb') as file:
file.write(image.content)
(You can specify the location of where you want the file to download by adding the PATH before the file name, like this:)
image = requests.get("https://i.imgur.com/5SMNGtv.png")
with open("C://Users//Admin//Desktop//image.png", 'wb') as file:
file.write(image.content)
Solution 1.
Access denied with the following error:
Cannot retrieve the public link of the file. You may need to change
the permission to 'Anyone with the link', or have had many accesses.
You may still be able to access the file from the browser:
https://drive.google.com/uc?id=<id>
In the sharing tab on gdrive (Right click on image, open Share or Get link), please change privacy to anyone with the link. Hopefully your code should work.
Solution 2.
If you can use Google Colab, then you can mount gdrive easily and access files there using
from google.colab import drive
drive.mount('/content/gdrive')
Google has this policy that they do not accept your regular google-/gmail-password. They only accept so called "App Passwords" that you need to create for your google-account in order to authenticate if you are using thirdparty apps
I'm working on a python script that would manage a file on a shared drive using Google Drive API v3. However, when I try to download or replace that file, I get the following error:
An error occurred: <HttpError 404 when requesting https://www.googleapis.com/upload/drive/v3/files/1HhCxshcR17I4rM0gtBIvHzEH_jzd7nV2?alt=json&uploadType=multipart returned "File not found: 1HhCxshcR17I4rM0gtBIvHzEH_jzd7nV2.">
Here is my code
def update_file(service, file_id, new_title, new_description, new_mime_type,
new_filename):
"""Update an existing file's metadata and content.
Args:
service: Drive API service instance.
file_id: ID of the file to update.
new_title: New title for the file.
new_description: New description for the file.
new_mime_type: New MIME type for the file.
new_filename: Filename of the new content to upload.
new_revision: Whether or not to create a new revision for this file.
Returns:
Updated file metadata if successful, None otherwise.
"""
try:
# File metadata
file_metadata = {
'name': new_title,
'description': new_description
}
# File's new content.
media_body = MediaFileUpload(
new_filename, mimetype=new_mime_type) # In this case the file is small so no resumable flag needed
# Send the request to the API.
updated_file = service.files().update(
fileId=file_id,
body=file_metadata,
media_body=media_body).execute()
return updated_file
except Exception as e:
print ('An error occurred: %s' % e)
return None
#Update an existing entry in database
def updateItem(dataIndex):
userInput = input(header[2]) #Update quantity
if userInput[0] == '+':
ws['C' + str(dataIndex)] = str( int(ws['C' + str(dataIndex)].value) + int(userInput[1:]) )
elif userInput[0] == '-':
ws['C' + str(dataIndex)] = str( int(ws['C' + str(dataIndex)].value) - int(userInput[1:]) )
else:
ws['C' + str(dataIndex)] = str( userInput )
ws['D' + str(dataIndex)] = today #Update date
wb.save(filepath)
How do I use Google Drive API on a shared drive or document? This works fine in my personal drive.
If you want to use the file of file_id in your shared Drive with files().update() in Drive API v3, how about the following modification?
Modified script:
updated_file = service.files().update(
fileId=file_id,
body=file_metadata,
media_body=media_body,
supportsAllDrives=True # <--- Added
).execute()
The default value of supportsAllDrives is false. So in this case, true is used for using the file in the shared Drive.
Reference:
Files: update
I have the following code which react not as I would like. I want to download the last file created/uploaded into the folder. The uploaded files are all CSV. I realized that it did not download the last one. So trying to understand, I realized that it detects the last file properly only if I open it once in the bowser. Then the code work and the file is detected. Any contribution would be appreciated.
DRIVE = discovery.build('drive', 'v3', http=creds.authorize(Http()))
folder_id = "xxxxxxxxxxxxxxxxxxxxxFW4-YY"
results = team_drive.DRIVE.files().list(pageSize=1, fields="files(createdTime,name,id)", orderBy="createdTime desc", q="'" + folder_id + "' in parents and mimeType = 'application/vnd.google-apps.spreadsheet'", supportsAllDrives=True, includeItemsFromAllDrives=True).execute()
items = results.get('files', [])
For example, if I read the first 1 files as per the code above. I will get:
{'files': [{'id': '1-dI4MOuqknn9-R324cTI658EhVayM98niNlrxMblvoU', 'name': 'Balance-2020-07-11', 'createdTime': '2020-07-11T02:46:04.180Z'}
But there is a file which more recently created as of July 12th. It is not detected...
I manage to get the correct output with the following code:
results = team_drive.DRIVE.files().list(pageSize=1, orderBy="createdTime desc", q="'" + folder_id + "' in parents and mimeType = 'text/csv'", supportsAllDrives=True, includeItemsFromAllDrives=True).exec$
print(results)
The flow of my desired program is:
Upload an xlsx spreadsheet to drive (it was created using pandas to_excel)
Convert it to Google Sheets format
Specify that it is editable by anyone with the link
Get the link and share it with someone who will enter information
Download the completed sheet
I am currently using PyDrive, which solves steps 1 and 5, but there are a few unsolved problems.
How can I convert to google sheets format? I tried to just specify the mimeType as 'application/vnd.google-apps.spreadsheet' when I created the file to upload with PyDrive, but that gave me an error.
How can I set the file to be editable by anyone with the link? Once that is set, I can get the sharing link easily enough with PyDrive.
UPDATE: conversion from xlsx to google sheets is easy with a convert=True flag. See below. I am still seeking a way to set the sharing settings of my new file to "anyone with the link can edit".
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
gauth = GoogleAuth()
gauth.LocalWebserverAuth()
drive = GoogleDrive(gauth)
test_file = drive.CreateFile({'title': 'testfile.xlsx'})
test_file.SetContentFile('testfile.xlsx')
test_file.Upload({'convert': True})
There is an Optional query parameter of "convert", for both the "INSERT" and "COPY" method;
convert=true,
Whether to convert this file to the corresponding Google Docs format. (Default: false)
There is a python example here:
Google Documentation - Copy
You need to use the Python client library for the code to work.
from apiclient import errors
from apiclient.http import MediaFileUpload
# ...
def insert_file(service, title, description, parent_id, mime_type, filename):
"""Insert new file.
Args:
service: Drive API service instance.
title: Title of the file to insert, including the extension.
description: Description of the file to insert.
parent_id: Parent folder's ID.
mime_type: MIME type of the file to insert.
filename: Filename of the file to insert.
Returns:
Inserted file metadata if successful, None otherwise.
"""
media_body = MediaFileUpload(filename, mimetype=mime_type, resumable=True)
body = {
'title': title,
'description': description,
'mimeType': mime_type
}
# Set the parent folder.
if parent_id:
body['parents'] = [{'id': parent_id}]
try:
file = service.files().insert(
body=body,
convert=true,
media_body=media_body).execute()
# Uncomment the following line to print the File ID
# print 'File ID: %s' % file['id']
return file
except errors.HttpError, error:
print 'An error occured: %s' % error
return None
I haven't tried this, so you'll need to test it.
In order to set the file to be editable for anyone with the link , you have to insert a new permission with the following information:
from apiclient import errors
# ...
def share_with_anyone(service, file_id):
"""Shares the file with anyone with the link
Args:
service: Drive API service instance.
file_id: ID of the file to insert permission for.
Returns:
The inserted permission if successful, None otherwise.
"""
new_permission = {
'type': "anyone",
'role': "writer",
'withLink': True
}
try:
return service.permissions().insert(
fileId=file_id, body=new_permission).execute()
except errors.HttpError, error:
print 'An error occurred: %s' % error
return None
then to get the link you go to : file["alternateLink"]