【发布时间】:2021-06-28 15:33:17
【问题描述】:
我的模特:
class User(db.Model, UserMixin):
__tablename__ = 'api_user'
id = db.Column(db.Integer, primary_key=True)
first_name = db.Column(db.String(30), nullable=False, index=True)
last_name = db.Column(db.String(40), nullable=False, index=True)
email = db.Column(db.String(50), unique=True, nullable=False, index=True)
password = db.Column(db.String(50), nullable=False)
birth_date = db.Column(db.Date, nullable=False, index=True)
即使这是一个小错误(例如无效的电子邮件格式等),它也会抛出一个错误并跳过它会给这个新用户的 id。 表:
bulletin_board=# SELECT * FROM api_user;
id | first_name | last_name | email | password | birth_date
----+------------+-----------+-------------------+----------+------------
1 | Rejep | Mammedov | reppon@mail.ru | test123 | 1997-11-16
3 | Rejep | Mammedov | repposn@mail.ru | test123 | 1997-11-16
4 | Rejep | Mammedov | repposan@mail.ru | test123 | 1997-11-16
8 | Rejep | Mammedov | repposasn@mail.ru | test123 | 1997-11-16
(4 rows)
user = User(first_name='Rejep', last_name='Mammedov', email='repposasn@mail.ru', password='test123', birth_date='16-11-1997')
db.session.add(user)
db.commit()
【问题讨论】:
标签: database postgresql flask sqlalchemy flask-sqlalchemy