Google drive api python - 如何获取下载文档的可共享链接

Question

我正在从谷歌驱动器下载一些文档，然后我需要在本地处理数据，处理完数据后，我需要获取下载文档的可共享链接并将其附加到电子表格中，其中包含生成的处理数据，我的实际代码来了

import os
import pickle
import os.path
import io
import shutil

from lector_nombre import lectorNombre
from Tablas import lectorTablas
from googleapiclient.discovery import build
from google_auth_oauthlib.flow import InstalledAppFlow
from google.auth.transport.requests import Request
from googleapiclient.http import MediaIoBaseDownload

SCOPES = ['https://www.googleapis.com/auth/drive.file','https://www.googleapis.com/auth/drive']

def main():

    #----------------------Google drive auth-----------------------------
    creds = None
    # The file token.pickle stores the user's access and refresh tokens, and is
    # created automatically when the authorization flow completes for the first
    # time.
    if os.path.exists('token.pickle'):
        with open('token.pickle', 'rb') as token:
            creds = pickle.load(token)
    # If there are no (valid) credentials available, let the user log in.
    if not creds or not creds.valid:
        if creds and creds.expired and creds.refresh_token:
            creds.refresh(Request())
        else:
            flow = InstalledAppFlow.from_client_secrets_file(
                'credentials.json', SCOPES)
            creds = flow.run_local_server(port=0)
        # Save the credentials for the next run
        with open('token.pickle', 'wb') as token:
            pickle.dump(creds, token)

    # Call the Drive v3 API
    service = build('drive', 'v3', credentials=creds)

    #-----------------------Download the files---------------------
    
    # ID DE LA CARPETA A DESCARGAR
    query = "'1qQ3245SwqSAOeqsuBdsh-ZFCqm' in parents"

    response = service.files().list(q=query,
                                spaces='drive',
                                fields='files(id, name, parents)').execute()
    
    for document in response['files']:
        #file_id = service.files.list()
        request = service.files().get_media(fileId=document['id'])
        fileName = document['name']
        fh = io.BytesIO()
        downloader = MediaIoBaseDownload(fh, request)
        done = False
        while done is False:
            status, done = downloader.next_chunk()
            print(document['name'])
            print ("Download %d%%." % int(status.progress() * 100))
            print("--------------")
            fh.seek(0)
            with open(fileName, 'wb') as f:
                shutil.copyfileobj(fh, f, length=131072)
            origen = "./"+document['name']
            shutil.move(origen, "./storingDir")

------------------------------------------code to procces the downloaded data--------------------------



[...]


------------------------------------------getting the sharable link-------------------------------
[...]

我正在阅读documentation，但我不明白如何使用 webViewLink 获取链接

Answer 1

出于文档目的发布此内容。

正如您已经注意到的，在调用Files: get 时，默认情况下不会返回webViewLink：

files.get 方法可能只返回files 资源的id、name 和mimeType。

为了返回特定字段，您必须在请求的fields 参数中设置这些字段。

在这种情况下，它可能是这样的：

fields='files(id, name, parents, webViewLink)'

参考：

• Return specific fields for a file