Node.js 如果include不'；在sequelize中找不到任何关系数据

我试图找出一本书可以有多个横幅的书

所以我在找书的时候用了很多横幅

这是工作的罚款，但我希望反应是不同的，如果在案件书并没有任何横幅

Book.findAll({
                order: [
                    ['updated_at', 'ASC']
                ],
                attributes: {
                    exclude: ["thumbnail", "listenTime", "created_at", "today_book_day"],
                },
                include: [{
                        model: banner,
                        as: 'banners',
                        attributes: [
                            [banner_attribute, 'url']
                        ],
                        where: whereAttribute,
                        required: false
                    }
                ]
            }).then(books => {
                var totalPage = 1;
                if (totalBooks){
                    totalPage = Math.ceil(totalBooks / pagesize);
                }
                var metaInfo = {
                    currentPage: parseInt(page),
                    totalPage: totalPage,
                    pagesize: pagesize
                };
                var output = {
                    'meta': metaInfo,
                    "data": books,
                    "status": 200
                };
                res.send(output);
            });

得到如下响应：

{
    "meta": {
        "currentPage": 1,
        "totalPage": 3,
        "pagesize": 100
    },
    "data": [
        {
            "id": 114,
            "rating": "5.00",
            "like": -1,
            "comment": 0,
            "listen": 49,
            "category": "[]",
            "order": 9,
            "updated_at": 1589475237,
            "thumbnailUrl": null,
            "banners": []
        },
        "status": 200
    ]
}

{
    "meta": {
        "currentPage": 1,
        "totalPage": 3,
        "pagesize": 100
    },
    "data": [
        {
            "id": 114,
            "rating": "5.00",
            "like": -1,
            "comment": 0,
            "listen": 49,
            "category": "[]",
            "order": 9,
            "updated_at": 1589475237,
            "thumbnailUrl": null,
            "banners": [
                {
                    "url" : null
                }
            ],
        },
        "status": 200
    ]
}

但我希望JSON响应如下所示：

{
    "meta": {
        "currentPage": 1,
        "totalPage": 3,
        "pagesize": 100
    },
    "data": [
        {
            "id": 114,
            "rating": "5.00",
            "like": -1,
            "comment": 0,
            "listen": 49,
            "category": "[]",
            "order": 9,
            "updated_at": 1589475237,
            "thumbnailUrl": null,
            "banners": []
        },
        "status": 200
    ]
}

{
    "meta": {
        "currentPage": 1,
        "totalPage": 3,
        "pagesize": 100
    },
    "data": [
        {
            "id": 114,
            "rating": "5.00",
            "like": -1,
            "comment": 0,
            "listen": 49,
            "category": "[]",
            "order": 9,
            "updated_at": 1589475237,
            "thumbnailUrl": null,
            "banners": [
                {
                    "url" : null
                }
            ],
        },
        "status": 200
    ]
}

这是无效的JSON

"banners": [url" : null] // table of... what?

我猜你想要这样的东西

"banners": [ {"url" : null} ] // table with one object, which has key 'url' and value 'null'

---

但若banners数组为空，那个么您可以断定url为空。您不需要显式地放置此信息。

好的，JSON无效，我已经修复了它。但是对于期望的响应，这是android开发者的要求，响应应该是这样的。如果是这样，那么我认为您必须添加一些逻辑（例如，在控制器中）以在发送响应之前更改该响应。