我正在使用聚合从子文档中分页数据.子文档没有严格的架构,因此每个文档的子文档可以有所不同,这使我难以构造输出,因为我不知道子文档的字段名称.

我在用什么

蒙戈3.0.0

节点0.10.33

猫鼬3.9.7

我的模特

var BodySchema = new Schema({random: String},{strict:false});

var FeedSchema = new Schema({
    name: String,
    body:[BodySchema]
});

我的数据看起来像这样

[{
    _id:"...",
    name:"Power Rangers feed",
    body:[
        {
            "_id":"...",
            "name" : "Jason",
            "colour" : "Red",
            "animal" : "T-rex"
        },
        {
            "_id":"...",
            "name" : "Billy",
            "colour" : "Blue",
            "animal" : "Triceratops"
        },
        {
            "_id":"...",
            "name" : "Zach",
            "colour" : "Black",
            "animal" : "Mastadon"
        }
    ]
},
{
    _id:"...",
    name:"Transformers feed",
    body:[
        {
            "_id":"...",
            "name" : "Optimus Prime",
            "team" : "Autobots",
            "class" : "leader"
            "alt-mode" : "truck"
        },
        {
            "_id":"...",
            "name" : "Bumblebee",
            "team" : "Autobots",
            "class" : "scout"
            "alt-mode" : "VW Beetle"
        },
        {
            "_id":"...",
            "name" : "Blaster",
            "team" : "Autobots",
            "class" : "Commmunicator"
            "alt-mode" : "Sterio"
        },
        {
            "_id":"...",
            "name" : "Hotrod",
            "team" : "Autobots",
            "class" : "Warrior"
            "alt-mode" : "Car"
        }
    ]
}]

我当前的汇总代码如下所示

feed.aggregate([
    {'$match':{_id:id('550234d3d06039d507d238d8')}},
    {'$unwind':'$body'},
    {'$skip':2},
    {'$limit':2},
    {
        '$project': {
            '_id':"$body._id",
            'body': "$body"
        }

    },
], function(err, result){

    if(err){return(res.send(500, err))}

    res.send(result);

});

我目前的结果看起来像这样

[{
    _id:"...",
    body:{
        "_id":"...",
        "name" : "Blaster",
        "team" : "Autobots",
        "class" : "Commmunicator"
        "alt-mode" : "Sterio"
    }
},
{
    _id:"...",
    body:{
        "_id":"...",
        "name" : "Hotrod",
        "team" : "Autobots",
        "class" : "Warrior"
        "alt-mode" : "Car"
    }
}]

我想要的结果看起来像这样

[
    {
        "_id":"...",
        "name" : "Blaster",
        "team" : "Autobots",
        "class" : "Commmunicator"
        "alt-mode" : "Sterio"
    },
    {
        "_id":"...",
        "name" : "Hotrod",
        "team" : "Autobots",
        "class" : "Warrior"
        "alt-mode" : "Car"
    }
]

我的问题

如何获得所需的结果结构.

解决方法:

因此,您将对此感到讨厌,但这就是聚合管道中的$project和$group阶段的工作方式.您需要“明确地”在输出中指定所需的所有字段.因此：

feed.aggregate([
    {'$match':{_id:id('550234d3d06039d507d238d8')}},
    {'$unwind':'$body'},
    {'$skip':2},
    {'$limit':2},
    {
        '$project': {
            '_id':"$body._id",
            'name': '$body.name',
            'team': '$body.team',
            'class': '$body.alt-mode'
        }
    },
], function(err, result){

那确实是唯一的方法.

但是,确实有很强的理由,并且大多数原理都在SQL SELECT中进行了备份,适用时*除外.

一般前提是,这类似于一般操作中的unix管道,因此,如果您认为：

grep | awk | sed

然后,这些操作在结构上似乎更加合乎逻辑.