mysql - 使用sequelize，在where子句中使用连接表的列值的正确方法是什么？

Question

我正在MySQL 数据库上使用sequelize。我正在获取一张表上的所有记录，同时将其与另一张表连接起来。如果不满足连接表上的 WHERE 条件，我尝试不包含结果。在 WHERE 子句中，我尝试将主表中的值与连接表进行比较，但它不起作用:

const expiredMemberships = await Membership.findAll({
include: [
  {
    model: User,
    attributes: ['id'],
    include: [
      {
        model: Post,
        required: true,
        attributes: ['id', 'created'],
        where: {
          created: {
            [Op.gt]: '$Membership.created$',
          },
        },
      },
    ],
  },
],
attributes: ['id', 'created'],
where: {
  isActive: true,
},

});

它创建以下查询:

SELECT `Membership`.`id`, 
       `Membership`.`created`, 
       `user`.`id` AS `user.id`, 
       `user->posts`.`id` AS `user.posts.id`, 
       `user->posts`.`created` AS `user.posts.created` 
  FROM `membership` AS `Membership` 
  LEFT OUTER JOIN ( `user` AS `user` INNER JOIN `post` AS `user->posts` 
       ON `user`.`id` = `user->posts`.`user_id` 
      AND `user->posts`.`created` > 'Invalid date' ) 
  ON `Membership`.`user_id` = `user`.`id` 
  AND (`user`.`deleted` = '9999-12-31 23:59:59') 
WHERE `Membership`.`is_active` = true;

您可以看到它显示'$Membership.created$'为无效日期。在 where 子句中使用连接表的列值的正确方法是什么？

Answer 1

您可以使用别名为成员(member)表的创建字段执行类似的操作，例如membership_created

const expiredMemberships = await Membership.findAll({
   include: [
    {
      model: User,
      attributes: ['id'],
      include: [
         {
             model: Post,
             required: true,
             attributes: ['id', 'created'],
             where: {
             created: {
                [Op.gt]: 'membership_created',
              },
         },
     },
   ],
  },
],
attributes: ['id', ['created', 'membership_created']],
where: {
      isActive: true,
    },
});