【发布时间】:2023-03-10 01:16:01
【问题描述】:
我可以使用Role.id == 4将用户列为审阅者,并且可以由当前用户选择:
def reviewer_choices():
return User.query.join(User.roles).filter(Role.id == 4)
form_extra_fields = {
'reviewer1': sqla.fields.QuerySelectField(
label='Reviewer1',
query_factory=reviewer1_choices,
)}
现在,如何使用Role.id == 4 和Team.id ==
current_user.teams.id 查询用户? (限制审阅者和当前用户属于同一团队)
我在下面尝试过,但没有成功:
User.query.join(User.roles).join(User.teams).
filter(Role.id == 4).filter(Team.id = current_user.teams.id)
类定义如下,谢谢:
class Role(db.Model, RoleMixin):
id = db.Column(db.Integer(), primary_key=True)
name = db.Column(db.String(80), unique=True)
def __str__(self):
return self.name
class Team(db.Model, RoleMixin):
id = db.Column(db.Integer(), primary_key=True)
name = db.Column(db.String(80), unique=True)
description = db.Column(db.String(255))
def __str__(self):
return self.name
class User(db.Model, UserMixin):
id = db.Column(db.Integer, primary_key=True)
roles = db.relationship('Role', secondary=roles_users,
backref=db.backref('users', lazy='dynamic'))
teams = db.relationship('Team', secondary=teams_users,uselist=False,
backref=db.backref('users', lazy='dynamic'))
email = db.Column(db.String(255), unique=True)
def __str__(self):
return self.email
class Project(db.Model):
id = db.Column(db.Integer, primary_key=True)
reviewer = db.Column(db.Unicode(128))
def __unicode__(self):
return self.name
【问题讨论】:
标签: python flask flask-sqlalchemy flask-admin