Question

我有两个表，sensors和readings。我需要在sensors中列出readings中最新值的所有条目。 readings表有一个指向sensors的外键。我如何用Sequelize实现这一目标？

以下是应该发生的事情：

sensors:

---------------
| id | name   |
---------------
| 1  | Test   |
---------------
| 2  | Test 2 |
---------------

readings:

---------------------------------------------
| id | sensor | value | time                |
---------------------------------------------
| 1  | 2      | 25.5  | 2016-06-14 17:01:50 |
---------------------------------------------
| 2  | 2      | 23.5  | 2016-06-14 17:04:50 |
---------------------------------------------
| 3  | 1      | 21.7  | 2016-06-14 17:06:00 |
---------------------------------------------

result:

---------------------------------------------
| id | name   | value | time                |
---------------------------------------------
| 1  | Test   | 21.7  | 2016-06-14 17:06:00 |
---------------------------------------------
| 2  | Test 2 | 23.5  | 2016-06-14 17:04:50 |
---------------------------------------------

Answer 1

我认为您可以使用sequelizejs的{{3}}来执行此操作;）

sequelize.query(
    'select s.*, r.value, r.time' +
    'from sensors s ' +
    'left join readings r on s.id = r.sensor' +
    'and (r.sensor, r.time) in (' +
    'select sensor, max(time) from readings group by sensor' +
    ')', 
    {type : sequelize.QueryTypes.SELECT})
.then(function(result) {
        console.log(result);
});

Answer 2

假设您正确建模关系，您应该能够执行如下查询：

models.sensors.findAll({
    include: [{
        model: models.readings,
        order: [['time', 'DESC']],
        limit: 1
    }],
    order: [['id', 'ASC']]
}).then(results => results.map(result => {
    return {
        id: result.id,
        name: result.name,
        value: result.readings[0].value,
        time: result.readings[0].time
    }
})).then(results => {
    // The rest of your logics here...
})

从一个表中选择所有表并加入另一个表

2 个答案: