Question

我是Sequelize ORM的新手。我有表用户，角色和投票。

我想让一个用户的角色（可能很少）和投票表的平均费率作为评级字段。我遇到了group by问题。

const id = req.params.id;
user
.findOne({
  where: { id },
  attributes: [
    "id",
    "username",
    "email",
    "avatar",
    "createdAt",
  ],
  include: [
    { model: role, attributes: ["name"] },
    { model: vote, attributes: ["rate"] },
  ],
})
.then(user => {
  res.status(200).json({ user });
})
.catch(err => {
  console.log(err);
});

上面的代码获得下一个json

{
  "id": 2,
  "username": "John Connor",
  "email": "johny@gmail.com",
  "avatar": "/public/static/img/uploads/avatar.jpg",
  "createdAt": "2018-03-05T20:25:56.693Z",
  "role": {
    "name": "USER_ROLE"
  },
  "votes": [
    {
      "rate": 4
    },
    {
      "rate": 5
    },
    {
      "rate": 2
    }
  ]
}

我不知道如何正确地执行复杂查询并获得平均值而不是让所有领域率。你怎么做这样的查询？也许我应该单独做这个或者有点不同？

Answer 1

你走了：

user.findOne({
    where: { id },
    attributes: [
        "id",
        "username",
        "email",
        "avatar",
        "createdAt",
        [sequelize.fn('AVG', sequelize.col('votes.rate')) ,'average_vote'] // perform average function like this
    ],
    include: [
        { model: role, attributes: ["id","name"] },
        { model: vote, attributes: [] }, // remove all the attributes from vote model
    ],
    // group by by the rest 2 table's (change user. and role. as per table name or query)
    group : ['user.id','role.id']

})

注意：



请阅读代码

中的评论

可能存在表格名称问题，因此您需要检查一下，其余部分将100％正常工作。

Answer 2

试试这个：

const id = req.params.id;
user
.findOne({
  where: { id },
  attributes: [
    "id",
    "username",
    "email",
    "avatar",
    "createdAt",
  ],
  include: [
    { model: role, attributes: ["name"] },
    { model: vote, attributes: [[sequelize.fn('avg', sequelize.col('rate')), 'rate']] },
  ],
group: ['id', 'username', 'email', 'avatar', 'createdAt', 'name'],
})
.then(user => {
  res.status(200).json({ user });
})
.catch(err => {
  console.log(err);
});

如何在Sequlize中使用多个包含正确编写复杂查询？

2 个答案: