Node.js 如何使用mongoose populate在多个案例中仅返回一个客户端?
我正在使用mongoose填充来尝试在我的模式中创建一对多关系Node.js 如何使用mongoose populate在多个案例中仅返回一个客户端?,node.js,mongodb,express,mongoose,Node.js,Mongodb,Express,Mongoose,我正在使用mongoose填充来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: {
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 }
});
我希望能够查询数据库,以获取具有给定客户id的所有案例。
我有这个工作,但不是我想要的方式。现在,当我使用populate方法时
在我的路线中,我得到了所有带有客户id的案例,但我也得到了
所有客户端,即使所有客户端都完全相同。对于每个案例,返回相同的客户端似乎是浪费资源。有没有办法只返回客户一次,然后返回所有相关案例
app.get('/cases/:id', function( req, res ) {
Case
.find( { _client: req.params.id } )
.populate('_client')
.exec( function( err, cases ) {
res.send( cases );
});
});
以下是我得到的反馈:
[
{
"_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
}
]
这似乎不适合将所有这些膨胀发送到我的视图进行渲染。我是否应该像这样使用一对多填充?
除了嵌入式文档,我在mongoose.com上看到的所有示例都是一对一的。如果您不想在每个案例中重复客户端,最好分别查询客户端,然后将该结果与
案例
查询的结果相结合(无填充
)以你需要的任何方式
顺便说一句,
“/cases/:id”
是一个相当混乱的URL,用于按客户id而不是案例id获取案例。我意识到这似乎有点混乱。为了测试的目的,我很快就把它组装起来了。不过我会试试看。