mysql - 连接两个表时将 where 子句转换为 and 子句的 Sequelize

Question

我有两个表Coupons 和LocationCoupons。

Coupons
--------------
id

LocationCoupons
----------------
id
coupon_id
location_id

我想从 Coupons 中选择 LocationCoupons 中不存在的所有记录。
我可以通过编写如下所示的原始查询来获得所需的结果。

const [results, metadata] = await sequelize.query("
    SELECT c.id, c.coupon_code FROM coupons c 
    WHERE NOT EXISTS 
    (SELECT l.id FROM location_coupon_map l WHERE c.id = l.coupon_id)");

但是当我尝试对关联执行相同操作时，使用联接，传递的 where 子句正在转换 and。

const coupons = await Coupon.findAll({
            include: {
                model: LocationCoupon,
                where: {
                    coupon_id: null
                },
                required: false,
                attributes: []
            },
            attributes: ["id", "coupon_code"]
        })

以上关联正在生成以下查询。

SELECT `Coupon`.`id`, `Coupon`.`coupon_code` FROM `coupons` AS `Coupon` 
LEFT OUTER JOIN `location_coupon_map` AS `LocationCoupons` 
ON `Coupon`.`id` = `LocationCoupons`.`coupon_id` 
AND `LocationCoupons`.`coupon_id` IS NULL;

虽然应该如此。

SELECT `Coupon`.`id`, `Coupon`.`coupon_code` FROM `coupons` AS `Coupon` 
LEFT OUTER JOIN `location_coupon_map` AS `LocationCoupons` 
ON `Coupon`.`id` = `LocationCoupons`.`coupon_id` 
WHERE `LocationCoupons`.`coupon_id` IS NULL;

我在这里遗漏了什么吗？

Answer 1

我修好了。 Sequelize 文档不太好。他们应该更好地描述所有可能的关联查询。

 const coupons = await Coupon.findAll({
            where: {
                "$LocationCoupons.coupon_id$": null
            },
            include: {
                model: LocationCoupon,
                attributes: []
            },
            attributes: ["id", "coupon_code"]
        })

现在生成的查询和结果符合预期。

SELECT `Coupon`.`id`, `Coupon`.`coupon_code` FROM `coupons` AS `Coupon` 
LEFT OUTER JOIN `location_coupon_map` AS `LocationCoupons` 
ON `Coupon`.`id` = `LocationCoupons`.`coupon_id` 
WHERE `LocationCoupons`.`coupon_id` IS NULL;