在stackoverflow上似乎有很多关于这个主题的Q / A,但我似乎无法在任何地方找到确切的答案。
是)我有的:
我有公司和人员模型:
var mongoose = require('mongoose');
var PersonSchema = new mongoose.Schema{
name: String,
lastname: String};
// company has a reference to Person
var CompanySchema = new mongoose.Schema{
name: String,
founder: {type:Schema.ObjectId, ref:Person}};
我需要的:
找到所有姓氏为“Robertson”的人成立的公司
我尝试了什么:
Company.find({'founder.id': 'Robertson'}, function(err, companies){
console.log(companies); // getting an empty array
});
然后我认为Person不是嵌入式的,而是引用的,所以我使用populate来填充创始人Person,然后尝试使用find和'Robertson'lastname
// 1. retrieve all companies
// 2. populate their founders
// 3. find 'Robertson' lastname in populated Companies
Company.find({}).populate('founder')
.find({'founder.lastname': 'Robertson'})
.exec(function(err, companies) {
console.log(companies); // getting an empty array again
});
我仍然可以使用Person的id作为String来查询公司。但这并不是我想要的,因为你可以理解
Company.find({'founder': '525cf76f919dc8010f00000d'}, function(err, companies){
console.log(companies); // this works
});
您不能在单个查询中执行此操作,因为MongoDB不支持连接。相反,你必须将它分成几个步骤:
// Get the _ids of people with the last name of Robertson.
Person.find({lastname: 'Robertson'}, {_id: 1}, function(err, docs) {
// Map the docs into an array of just the _ids
var ids = docs.map(function(doc) { return doc._id; });
// Get the companies whose founders are in that set.
Company.find({founder: {$in: ids}}, function(err, docs) {
// docs contains your answer
});
});
如果有人在最近的时间遇到过这种情况,Mongoose现在支持使用名为Populate的功能加入类似功能。
从Mongoose文档:
Story.findOne({
title: 'Casino Royale'
}).populate('author').exec(function (err, story) {
if (err) return handleError(err);
console.log('The author is %s', story.author.name);
// prints "The author is Ian Fleming"
});