假设有一个User模型和一个Post模型。在这种情况下,用户的帖子会很多;用户将是父级,而邮政将是子级。是否可以直接查询帖子?
例如,如果我想做类似的事情
app.get('/post/search/:query', (req,res) => {
Posts.find({title: req.params.query }, (err,post) => {
res.send(JSON.stringify(post))
})
})
还是必须要做:
app.get('/post/search/:query',(req,res) => {
let resultsFromQuery = [];
User.find({'post.title':req.params.query'}, (err,user) => {
user.posts.forEach((post) => {
if(post.title === req.params.query){
resultsFromQuery.push(post);
}
})
})
res.send(JSON.stringify(resultsFromQuery))
})
编辑:这是我的架构。
用户架构(父)
const mongoose = require('mongoose'),
Schema = mongoose.Schema,
PostSchema = require('./post.js');
let UserSchema = new Schema({
username: {
type: String,
required: true,
unique: true
},
password: {
type: String,
required: true
},
posts: [PostSchema]
})
module.exports = mongoose.model('User',UserSchema);
发布架构(子级)
const mongoose = require('mongoose'),
Schema = mongoose.Schema;
let PostSchema = new Schema({
title: {
type: String
},
description: {
type: String
},
image: {
type: String
},
original_poster: {
id: {
type: String,
required: true
},
username: {
type: String,
required: true
}
},
tags: {
type: [String],
required: true
}
})
module.exports = PostSchema;
编辑:
这是一个样本文件
db.users.find({username:'john'})的结果
{
"_id" : ObjectId("5a163317bf92864245250cf4"),
"username" : "john",
"password" : "$2a$10$mvE.UNgvBZgOURAv28xyA.UdlJi4Zj9IX.OIiOCdp/HC.Cpkuq.ru",
"posts" : [
{
"_id" : ObjectId("5a17c32d54d6ef4987ea275b"),
"title" : "Dogs are cool",
"description" : "I like huskies",
"image" : "https://media1.giphy.com/media/EvRj5lfd8ctUY/giphy.gif",
"original_poster" : {
"id" : "5a163317bf92864245250cf4",
"username" : "john"
},
"tags" : [
"puppies",
"dogs"
]
}
],
"__v" : 1
}
最佳答案
是的,您可以直接从用户模型中找到帖子标题。像波纹管
User.find({"posts.title": "Cats are cool"}, (err, users) => {
if(err) {
// return error
}
return res.send(users)
})
这将返回具有所有帖子的用户,而不仅仅是匹配的帖子标题。因此,要返回仅匹配的帖子标题,可以使用
$
位置运算符。喜欢这个查询User.find({"posts.title": "Cats are cool"},
{username: 1, "posts.$": 1}, // add that you need to project
(err, users) => {
if(err) {
// return error
}
return res.send(users)
})
只返回匹配的帖子