使用 MongoDB 根据该记录的值获取不同的记录

Question

我有 collection 家餐厅记录。这 collection 中的一些餐厅属于某个集团（连锁餐厅，例如 KfC 等），而其他则没有任何集团（个人餐厅，不属于任何链）。

示例：

餐厅collection

{_id:"1",title:"rest1",address:"somethingx",chain_id:"123"},
{_id:"2",title:"rest2",address:"somethingx",chain_id:"123"},
{_id:"3",title:"rest3",address:"somethingy",chain_id:"0"},
{_id:"4",title:"rest4",address:"somethingx",chain_id:"0"}

链collection：

{_id:"123",chain_name:"VSWEETS",address_deatils:[
                                          {restID:"",address:"somethingx"},
                                          {restID:"",address:"somethingx"}
                                         ]
}

{_id:"456",chain_name:"KFC",address_deatils:[]}

我需要获取具有相似 chain_id 的独特餐厅，即，如果它属于某个连锁店，则只应出现单个餐厅(chain_id !=0)

Answer 1

您可以为此使用聚合框架。聚合管道的第一步是 $match operator that filters the restaurants on the address. The $group pipeline stage will then group the filtered documents by the chain_id key and applies the accumulator expression $first to the $$ROOT system variable on each group. You can the reshape the documents using the $project 管道阶段。

为您提供所需结果的最终聚合管道如下：

db.restaurant.aggregate([
    {
        "$match": { "address" : "somethingx" }
    },
    {
        "$group": {
            "_id": "$chain_id",
            "data": { "$first": "$$ROOT" }
        }
    },
    {
        "$project": {
            "_id" : "$data._id",
            "title" : "$data.title",
            "address" : "$data.address",
            "chain_id" : "$data.chain_id"
        }
    }
])

输出:

/* 0 */
{
    "result" : [ 
        {
            "_id" : "4",
            "title" : "rest4",
            "address" : "somethingx",
            "chain_id" : "0"
        }, 
        {
            "_id" : "1",
            "title" : "rest1",
            "address" : "somethingx",
            "chain_id" : "123"
        }
    ],
    "ok" : 1
}

使用 MongoDB 根据该记录的值获取不同的记录

Fetch distinct record based on values for that record with MongoDB

mongodb

nosql

sails.js

sails-mongo