Question

我有3个模型：Post，Photo和Audio。

帖子可以有照片或音频剪辑（或两者都没有）：

models.Post.hasOne(models.Photo, { as: 'photo' });
models.Post.hasOne(models.Audio, { as: 'audio' });

我正在尝试选择那些既没有 - 也就是文本帖子。我一直在挖掘，我到了（从here获取信息）：

db.Post.findAll({
    where: {
        '$photo.id$': { $eq: null },
        '$audio.id$': { $eq: null }
    },
    order: [
        ['createdAt', 'DESC']
    ],
    include: [{... "photo" and "audio" - both with required: true}],
}).then((posts) => {
    ...
});

然而，这似乎不起作用（我不认为我完全理解相关模型是如何被引用的）。此外，生成的查询使用INNER JOINS而不是LEFT JOINS，因此当它尝试加入Posts和Photos / Audios时，会生成一个空结果，而不是Post的所有值以及Photo和Audio的空列的结果。

生成的查询是：

SELECT ... FROM (
    SELECT *
    FROM "Posts" AS "Post"
    INNER JOIN "Photos" AS "photo"
        ON "Post"."id" = "photo"."postId"
    INNER JOIN "Audios" AS "audio"
        ON "Post"."id" = "audio"."postId"
    WHERE 
        "photo"."id" IS NULL AND
        "audio"."id" IS NULL
) AS ...

查询

SELECT *
FROM "Posts" AS "Post"
LEFT JOIN "Photos" AS "photo"
    ON "Post"."id" = "photo"."postId"
LEFT JOIN "Audios" AS "audio"
    ON "Post"."id" = "audio"."postId"
WHERE 
    "photo"."id" IS NULL AND
    "audio"."id" IS NULL

生成我正在寻找的结果集。我认为我的问题是每个包含模型的include: [{ ... required: true }, ...]标志 - 但我不确定如何将包含在中以查询但不包含要求它的结果采集。

在我的梦想世界中，续集查询看起来像

db.Post.findAll({ where: { photo: { $eq: null }, audio: ... } })

Answer 1

使用required: false生成左外部联接。

const lonelyPosts = await db.Post.findAll({
  include: [
    { model: db.Photo, required: false, attributes: [] },
    { model: db.Audio, required: false, attributes: [] },
  ],
  where: [
    Sequelize.where(Sequelize.col('photo.id'), null), 
    Sequelize.where(Sequelize.col('audio.id'), null),
  ],
})

Answer 2

尝试不包含

db.Post.findAll({
        where: {
            '$photo.id$': { $eq: null },
            '$audio.id$': { $eq: null }
        },
        order: [
            ['createdAt', 'DESC']
        ]
    }).then((posts) => {
        ...
    });

续集关联模型是空的

2 个答案: