在 Virtual Populate 中,如何定义 foreignField?
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【中文标题】在 Virtual Populate 中,如何定义 foreignField?【英文标题】:In Virtual Populate, how do you define foreignField? 【发布时间】:2020-07-08 04:35:52 【问题描述】:考虑下面的代码:
require("./connection");
// //----------------------------------------------------
const mongoose = require("mongoose");
const Schema = mongoose.Schema;
const PersonSchema = new Schema(
name: String,
band: String,
father: String
);
const ManagerSchema = new Schema(
name: String,
country: String
);
const BandSchema = new Schema(
name: String
);
BandSchema.virtual("members",
ref: "Person", // The model to use
localField: "name", // Find people where `localField`
foreignField: "band", // is equal to `foreignField`
// If `justOne` is true, 'members' will be a single doc as opposed to
// an array. `justOne` is false by default.
justOne: false,
options: sort: name: -1 , limit: 5
);
BandSchema.virtual("managers",
ref: "Manager", // The model to use
localField: "name", // Find people where `localField`
foreignField: "country", // is equal to `foreignField`
// If `justOne` is true, 'members' will be a single doc as opposed to
// an array. `justOne` is false by default.
justOne: false,
options: sort: name: 1 , limit: 5
);
//BandSchema.set("toObject", virtuals: true );
BandSchema.set("toJSON", virtuals: true );
const Person = mongoose.model("Person", PersonSchema);
const Manager = mongoose.model("Manager", ManagerSchema);
const Band = mongoose.model("Band", BandSchema);
/**
* Suppose you have 2 bands: "Guns N' Roses" and "Motley Crue"
* And 4 people: "Axl Rose" and "Slash" with "Guns N' Roses", and
* "Vince Neil" and "Nikki Sixx" with "Motley Crue"
*/
// Person.create([
//
// name: "Axl Rose",
// band: "Guns N' Roses"
// ,
//
// name: "Slash",
// band: "Guns N' Roses"
// ,
//
// name: "Vince Neil",
// band: "Motley Crue"
// ,
//
// name: "Nikki Sixx",
// band: "Motley Crue"
//
// ]);
// Manager.create([
//
// name: "Bibi",
// country: "South Africa"
// ,
//
// name: "Storm",
// country: "Italy"
// ,
//
// name: "Wolverine",
// country: "Canada"
// ,
//
// name: "Jorge Pires",
// country: "Brazil"
//
// ]);
// Band.create([ name: "Motley Crue" , name: "Guns N' Roses" ]);
/////////////////////////////////////////////////////////////////////////
const app = require("express")();
app.use("/", (req, res) =>
Band.find()
.populate("members")
.populate("managers")
.exec(function(error, bands)
/* `bands.members` is now an array of instances of `Person` */
console.log(bands);
res.json(bands);
);
);
app.listen(3000, () =>
console.log("We are on port 3000");
);
/**
*https://***.com/questions/43882577/mongoosejs-virtual-populate
https://***.com/questions/60875380/populate-virtuals-does-not-seem-to-work-could-anyone-show-me-the-error
*/
考虑相关问题:
Mongoosejs virtual populate
Populate Virtuals does not seem to work. Could anyone show me the error?
我的问题是:你如何定义foreignField
?
Members
正确填充,但 manager 没有。
我知道问题出在foreignField
,因为如果我重复来自成员的所有信息,它将正确填充,但现在我们拥有成员并使用相同的数据源进行管理。
【问题讨论】:
【参考方案1】:经过一些研究,尝试在 Stack Overflow 上回答另一个问题 (mongoose: populate in mongoose which doesn't have any ObjectId ),我意识到它是如何工作的!
考虑部分代码:
BandSchema.virtual("members",
ref: "Person", // The model to use
localField: "name", // Find people where `localField`
foreignField: "band", // this field here has to match the ref path we want to populate!
justOne: false,
);
因此,诀窍是确保 foreignField
与 ref
模型中的字段匹配,localField
是您可以在填充模型中找到的字段名称:我们必须在 @ 之间进行匹配987654328@ 和localField
,更准确地说:localField
的值必须与实际情况下的foreignField
之一匹配,在数据库中,而不是在模式命名过程中。这就是猫鼬可以找到和填充的方式!
现在我意识到我遇到的困难是Virtual
在某种程度上与populate
的操作方向相反。您可以通过像树一样工作来进行图片填充:它只是从id
填充到文档中,而virtual
将填充不包含密钥的文档,包含密钥的文档就是要添加的文档填充过程:它以某种方式落后,这对我来说太难掌握了!要填充的文档没有刚刚填充的字段!令人兴奋!
结论和最后的评论
尽管如此,与populate
相比,我仍然觉得它有限。您仍然必须保留本地关键轨道,这并不能解决例如故事记忆问题,而且根据我的研究它更加有限。我看到的唯一优点是您不需要使用_id
,您可以使用任何您喜欢的密钥。我希望在这里使用How to save an JSON file using GridFs 来解决我的问题,但是由于我们仍然必须存储本地密钥,所以我落入了同样的陷阱!
更正
我很高兴我错了! Virtuals 太棒了!并在How to save an JSON file using GridFs 上解决了我的问题,我将在那里更新!
【讨论】:
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