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我正在使用 mongoose populate 尝试在我的模式中创建一对多关系

var clientSchema = new mongoose.Schema({
    name:  { type: String, required: true },
    title: { type: String, default: "N/S" },
    birthDate: { type: Date },
    ssn: { type: String, default: "N/S" },
    spouse: { type: String, default: "N/S" },
    notes: { type: String, default: "N/S" },
    address: { type: String, default: "N/S" },
    city: { type: String, default: "N/S" },
    state: { type: String, default: "N/S" },
    zip: { type: String, default: "N/S" },
    homePhone: { type: String, default: "N/S" },
    workPhone: { type: String, default: "N/S" },
    mobilePhone: { type: String, default: "N/S" },
    fax: { type: String, default: "N/S" },
    email: { type: String, default: "N/S" }
});
var caseSchema = mongoose.Schema({
    _client: { type: mongoose.Schema.Types.ObjectId, ref: 'Client' },
    name: { type: String, required: true },
    lead: { type: String },
    priority: { type: String },
    dateOpened: { type: Date },
    dateAccident: { type: Date },
    status: { type: String },
    sol: { type: Date },
    description: { type: String }
});

我想要做的是查询数据库以获取具有给定 _client id 的所有案例。我有这个工作，但不是我想要的方式。截至目前，当我在我的路线中使用填充方法时，我得到了具有该客户端 ID 的所有案例，但我也得到了所有客户端，即使所有客户端都完全相同. 为每种情况返回相同的客户端似乎是一种资源浪费。有没有办法只返回一次客户，然后所有相关的案例都用它？

app.get('/cases/:id', function( req, res ) {
    Case
    .find( { _client: req.params.id } )
    .populate('_client')
    .exec( function( err, cases ) {
        res.send( cases );
    });
});

这是我要回来的：

[
    {
        "_client": {
            "name": "John Doe",
            "birthDate": null,
            "_id": "51705a7ed0ecd0a906000001",
            "__v": 0,
            "email": "",
            "fax": "",
            "mobilePhone": "",
            "workPhone": "",
            "homePhone": "",
            "zip": "",
            "state": "",
            "city": "",
            "address": "",
            "notes": "",
            "spouse": "",
            "ssn": "",
            "title": "Mr"
        },
        "name": "test",
        "lead": "",
        "priority": "",
        "dateOpened": null,
        "dateAccident": null,
        "status": "",
        "sol": null,
        "description": "",
        "_id": "5170679df8ee8dd615000001",
        "__v": 0
    },
    {
        "_client": {
            "name": "John Doe",
            "birthDate": null,
            "_id": "51705a7ed0ecd0a906000001",
            "__v": 0,
            "email": "",
            "fax": "",
            "mobilePhone": "",
            "workPhone": "",
            "homePhone": "",
            "zip": "",
            "state": "",
            "city": "",
            "address": "",
            "notes": "",
            "spouse": "",
            "ssn": "",
            "title": "Mr"
        },
        "name": "newest case",
        "lead": "",
        "priority": "",
        "dateOpened": null,
        "dateAccident": null,
        "status": "",
        "sol": null,
        "description": "",
        "_id": "517067d8806f060b16000001",
        "__v": 0
    },
    {
        "_client": {
            "name": "John Doe",
            "birthDate": null,
            "_id": "51705a7ed0ecd0a906000001",
            "__v": 0,
            "email": "",
            "fax": "",
            "mobilePhone": "",
            "workPhone": "",
            "homePhone": "",
            "zip": "",
            "state": "",
            "city": "",
            "address": "",
            "notes": "",
            "spouse": "",
            "ssn": "",
            "title": "Mr"
        },
        "name": "adding new case",
        "lead": "Me",
        "priority": "Urgent",
        "dateOpened": null,
        "dateAccident": null,
        "status": "",
        "sol": null,
        "description": "",
        "_id": "51709a16806f060b16000002",
        "__v": 0
    }
]

将所有这些臃肿发送到我的视图进行渲染似乎并不正确。我什至应该像这样使用填充一对多吗？我在 mongoose.com 上看到的所有示例都是一对一的，嵌入式文档除外。

4

1 回答 1

2

如果您不希望客户端为每种情况重复，您最好单独查询客户端，然后以您需要的任何方式将该结果与Case查询结果（不带）组合。populate

顺便说一句，'/cases/:id'这是一个非常令人困惑的 URL，用于通过客户端 ID 而不是案例 ID 获取案例。

于 2013-04-19T02:06:57.690 回答