Sequelize:查找没有关联的条目

Fra*_*ger 3 sequelize.js

给出以下简单模型:

class A extends Model {}
A.init({
    aField: DataTypes.STRING,
}, { sequelize });

class B extends Model {}
B.init({
    bField: DataTypes.STRING,
}, { sequelize });

A.hasMany(B); // this creates the B.AId column


// populate

await A.create({
    aField: 'fooA',
    Bs: [{ bField: 'fooB' }]
}, { include: [B] });

await A.create({ aField: 'barA' });
await A.create({ aField: 'bazA' });
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我不知道如何选择第一个没有 B 条目的 A 模型实例。
前一种情况,仅返回barAbazA

以下不起作用,它返回foo

await A.findOne({
    include: [{
        model: B,
        required: false,
        where: {
            'AId': null,
        }
    }],
});
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生成的查询是:SELECT A.*, Bs.id AS Bs.id, Bs.bField AS Bs.bField, Bs.AId AS Bs.AId FROM (SELECT A.id, A.aField FROM As AS A LIMIT 1) AS A LEFT OUTER JOIN Bs AS Bs ON A.id = Bs.AId AND Bs.AId IS NULL

Emm*_*mma 7

这是一个极限和关联的问题。类似问题:https://github.com/sequelize/sequelize/issues/7585

Sequelize 将关联转变为子查询,但在您的用例中,您需要一个关联,然后限制总体关联结果。为此,您可以使用subQuery: false禁用子查询。

await A.findOne({
    include: [{
        model: B,
        required: false,
        where: {
            'AId': null,
        }
    }],
    subQuery: false
});
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这仍然无法实现你想要的,因为whereininclude会生成AND的条件ON

生成的查询:

SELECT `A`.`id`, `A`.`aField`, `A`.`createdAt`, `A`.`updatedAt`, `Bs`.`id` AS `Bs.id`, 
    `Bs`.`bField` AS `Bs.bField`, `Bs`.`createdAt` AS `Bs.createdAt`, 
    `Bs`.`updatedAt` AS `Bs.updatedAt`, `Bs`.`AId` AS `Bs.AId` 
FROM `As` AS `A` 
LEFT OUTER JOIN `Bs` AS `Bs` ON `A`.`id` = `Bs`.`AId` AND `Bs`.`AId` IS NULL 
LIMIT 1;
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你想要的是

...
... ON `A`.`id` = `Bs`.`AId` WHERE `Bs`.`AId` IS NULL 
LIMIT 1;
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为了得到这个WHERE,你可以添加where子句 for A

await A.findOne({
    include: [{
        model: B,
        required: false
    }],
    where: {
        '$Bs.AId$': null
    },
    subQuery: false
});
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