在尝试创建一个已经是用户的新用户时,使用Firebase身份验证的pyrebase包装器,pyrebase会将google API响应包装在HTTPError消息中。但是,当我尝试捕获此异常时,它不会将HTTPError识别为异常。我可以通过使用下面更详细的expect Exception as e显示来访问该异常。
代码:
config = {
"apiKey": os.environ.get('WEB_API_KEY'),
"authDomain": "project.firebaseapp.com",
"databaseURL": "https://project.firebaseio.com",
"storageBucket": "project.appspot.com",
"serviceAccount": os.environ.get('FIREBASE_APPLICATION_CREDENTIALS')
}
firebase = pyrebase.initialize_app(config)
auth = firebase.auth()
# Attempt to register a user that already exists
try:
user = auth.create_user_with_email_and_password('myemail@email.com', 'mypassword')
except HTTPError as e:
print('Handling HTTPError:', e)这将输出:
Traceback (most recent call last):
File "<console>", line 3, in <module>
NameError: name 'HTTPError' is not defined如果我采取更通用的方法并使用以下命令,我可以捕获错误:
try:
user = auth.create_user_with_email_and_password('myemail@email.com', 'mypassword')
except Exception as e:
print(e.args)然后,这将优雅地打印异常:
(HTTPError('400 Client Error: Bad Request for url: https://www.googleapis.com/identitytoolkit/v3/relyingparty/signupNewUser?key=<WEB_API_KEY>'), '{\n "error": {\n "code": 400,\n "message": "EMAIL_EXISTS",\n "errors": [\n {\n "message": "EMAIL_EXISTS",\n "domain": "global",\n "reason": "invalid"\n }\n ]\n }\n}\n')这给了我信息,但它是一个字符串。
如何访问异常消息中显示的响应JSON?
谢谢!
发布于 2020-08-14 03:17:25
json.loads(e.args[1])['error']['message']这将为您提供结果:EMAIL_EXISTS
https://stackoverflow.com/questions/61627506
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