afr*_*afr 3 python flask flask-sqlalchemy graphql graphene-python
我是 GraphQL 的相对新手,我正在尝试进行这样的查询
{
user(username: "Jon") {
name
last_lame
username
posts(in_draft : true) {
title
text
in_draft
update_at
}
}
}
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我想过滤用户在草稿中的帖子列表
我可以进行查询的唯一方法是通过模型的关系,但无法过滤草稿中的帖子。一对多
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
name = Column(String)
last_lame = Column(String)
username = Column(String)
class Post(Base):
__tablename__ = 'post'
id = Column(Integer, primary_key=True)
title = Column(String)
text = Column(String)
in_draft = Column(Boolean)
post_id = Column(Integer, ForeignKey('user.id'))
posts = relationship("User", backref='posts')
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有了这种关系,我用“backref = 'posts'”显示了帖子节点
我的对象:
class User(SQLAlchemyObjectType):
"""User Object."""
class Meta:
model = UserModel
interfaces = (relay.Node, )
class Post(SQLAlchemyObjectType):
"""Post Object."""
class Meta:
model = PostModel
# interfaces = (relay.Node, )
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询问:
class Query(graphene.ObjectType):
user = graphene.Field(lambda: User, username=graphene.String())
def resolve_user(self, info, username):
query = User.get_query(info)
return query.filter(UserModel.username == username).first()
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我想让帖子的查询属于用户
posts = graphene.List(lambda: Post, in_draft=graphene.Boolean())
def resolve_posts(self, info, in_draft):
query = Post.get_query(info)
return query.filter(PostModel.in_draft == in_draft).all()
schema = graphene.Schema(
query=Query,
types=[User, Post])
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有什么想法或建议吗?
老实说,鉴于已经定义了User和之间的关系,Post我将创建一个解析器来让这些组合结果同时传递两个参数,如下所示:
class Query(graphene.ObjectType):
filter_user_posts = graphene.List(
lambda: User,
username=graphene.String,
in_draft=graphene.Boolean,
)
def resolve_filter_user_posts(
self, info, username, in_draft
):
query = User.get_query(info=info)
query = query.join(User.posts)
query = query.filter(User.username == username)
query = query.filter(Post.in_draft == in_draft)
objs = query.all()
return objs
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然后您可以像这样简单地查询:
{
filterUserPosts(username: "Jon", in_draft: true) {
name
last_lame
username
posts{
title
text
in_draft
update_at
}
}
}
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