firebase 连接到数据库抛出无效数据库 URL
firebase connect to database throwing invalid Database URL
我有一个简单的 python 应用程序,我正在尝试连接到 Firebase,我下载了我的 cert.json 文件并拥有一个有效的证书来进行身份验证,但是当我尝试创建对根据文档的数据库,它抛出以下错误:
File "C:\Python3\lib\site-packages\firebase_admin\db.py", line 69, in reference
client = service.get_client(url)
File "C:\Python3\lib\site-packages\firebase_admin\db.py", line 798, in get_client
raise ValueError(
ValueError: Invalid database URL: "None". Database URL must be a non-empty URL string.
我在下面添加了我的代码:
import firebase_admin
import os
from firebase_admin import db
from firebase_admin import credentials
certfile = "firebase_cert.json"
credentials = credentials.Certificate(certfile)
database = firebase_admin.initialize_app(credentials)
ref = db.reference("/")
print(ref)
如 initializing the Admin SDK in Python 的 Firebase 文档所示,您需要在应用程序代码中为文档指定 URL:
# Fetch the service account key JSON file contents
cred = credentials.Certificate('path/to/serviceAccountKey.json')
# Initialize the app with a service account, granting admin privileges
firebase_admin.initialize_app(cred, {
'databaseURL': 'https://databaseName.firebaseio.com' //
})
ref = db.reference("/")
我有一个简单的 python 应用程序,我正在尝试连接到 Firebase,我下载了我的 cert.json 文件并拥有一个有效的证书来进行身份验证,但是当我尝试创建对根据文档的数据库,它抛出以下错误:
File "C:\Python3\lib\site-packages\firebase_admin\db.py", line 69, in reference client = service.get_client(url) File "C:\Python3\lib\site-packages\firebase_admin\db.py", line 798, in get_client raise ValueError( ValueError: Invalid database URL: "None". Database URL must be a non-empty URL string.
我在下面添加了我的代码:
import firebase_admin
import os
from firebase_admin import db
from firebase_admin import credentials
certfile = "firebase_cert.json"
credentials = credentials.Certificate(certfile)
database = firebase_admin.initialize_app(credentials)
ref = db.reference("/")
print(ref)
如 initializing the Admin SDK in Python 的 Firebase 文档所示,您需要在应用程序代码中为文档指定 URL:
# Fetch the service account key JSON file contents
cred = credentials.Certificate('path/to/serviceAccountKey.json')
# Initialize the app with a service account, granting admin privileges
firebase_admin.initialize_app(cred, {
'databaseURL': 'https://databaseName.firebaseio.com' //
})
ref = db.reference("/")