本文介绍了MongoDB聚合项目返回_id数组的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我们知道是否要获取_id数组,我们可以这样做:
As we know if we want to get an array of _id we can do:
db.collections.distinct("_id");我的问题是,如果我需要对聚合进行复杂的逻辑运算,该如何获取_id数组. 例如:
My question is how can I get an array of _id if I need to do a complicate logic with aggregate. Ex:
db.getCollection('users').aggregate({ $match : { is_register_completed : { $ne : true} } } //other operator like $lookup, $group , { $project : {_id:1} } )我知道
{ "_id" : "1", "_id" : "2" }我想要的就像我们做的独特
what i want is just like we do distinct
{[1,2]}已更新: 这就是我尝试使用$ group
Updated: this is what i try to do with $group
db.getCollection('users').aggregate({ $match : { is_register_completed : { $ne : true} } }, { $group: { _id:null, all:{$addToSet: "$_id"} } }, {$project: {_id:0,all:1}} )但我仍然得到
{ all : ["1","2"] }或者我可以在获得
{ "_id" : "1", "_id" : "2" },但是地图是客户端转换,我认为这会影响性能.
, but the map is client side transformation which I think it will affect the performance.
推荐答案引用自:如何通过mongodb聚合返回字符串数组
.aggregate()方法始终返回对象,无论您做什么 并不能改变.
The .aggregate() method always returns Objects no matter what you do and that cannot change.
原始答案: 尝试使用聚合框架:
Original Answer: Try the aggregation framework:
db.myCol.aggregate([ { $group:{_id:null, array:{$push:"$_id"}} }, { $project:{array:true,_id:false} } ])更多推荐
MongoDB聚合项目返回
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