mysql - 同表 Sequelize 上的内连接

Question

我已经编写了一个 SQL 查询，并尝试将其调整为 Sequelize，但无法弄清楚如何在同一个表上创建关联。这是 SQL:

SELECT t1.*
FROM `configurations` t1
INNER JOIN
(
    SELECT max(configuration_start) maxConfigStart, pb_id
    FROM `configurations`
    WHERE
        active = 1
        AND DATE(configuration_start) <= DATE(NOW())
    GROUP BY `pb_id`
) t2
  ON t1.`pb_id` = t2.`pb_id`
  AND t1.`configuration_start` = t2.`maxConfigStart`
WHERE t1.`active` = 1
ORDER BY t1.`configuration_start` DESC

这是模型定义:

configurations = sequelize.define 'configurations', {
  id:
    type: Sequelize.INTEGER
    primaryKey: true
    autoIncrement: true
  configuration_name: Sequelize.STRING
  configuration_start: Sequelize.DATE
  configuration_end: Sequelize.DATE
  created: Sequelize.DATE
  updated: Sequelize.DATE
}, {
  timestamps: false,
  scopes: {
    currentConfigTimes: {
      attributes: ['pb_id', [sequelize.fn('max', sequelize.col('configuration_start')), 'configuration_start']]
      where:
        active: 1
        configuration_start:
          lte: new Date()
      group: ['pb_id']
    },
    currentConfigIds: {
      attributes: ['pb_id', 'configuration_start', 'id']
      where:
        active: 1
        configuration_start:
          lte: new Date()
      order: [['configuration_start', 'DESC']]
    }
  }
}

如果有人能提供任何帮助，我们将不胜感激。本质上，我试图获取每个 pb_id 的最新配置。也许有更好的方法？

Answer 1

目前看来，使用 Sequelize 实际上无法实现这一点。在花费了大量时间研究他们的文档并尝试了解执行此操作的最佳方法之后，我发现了 this Github issue试图解决完全相同的问题。此时最好的解决方案就是使用原始查询。