从嵌套数组集合中检索的mongoDB查询 [英] mongoDB query for retrieving from nested array collection

问题描述

{
    "_id" : ObjectId("576155a6cd87b68f7e6e42c9"),
    "First_Name" : "ok",
    "Last_Name" : "jaao",
    "Email" : "[e-mail]",
    "Sessions" : [
        {
            "Last_Login" : "Wed, Jun 14, 2016 6:48 PM",
            "Class" : "fb",
            "ID" : "123"
        },
        {
            "Last_Login" : "Wed, Jun 15, 2016 6:48 PM",
            "ID" : "111",
            "Class" : "fb"
        }
    ],
    "Count" : 2
},
{
    "_id" : ObjectId("576155ccf6d8979e7e77df27"),
    "First_Name" : "abc",
    "Last_Name" : "xyz",
    "Email" : "[e-mail]",
    "Sessions" : [
        {
            "Last_Login" : "Wed, Jun 15, 2016 6:49 PM",
            "Class" : "fb",
            "ID" : "123"
        }
    ],
    "Count" : 1
}

这是我的json结构.我想要一个mongoDB查询来检索今天登录的每个用户，即Last_Login具有今天的日期.

This is my json structure. and I want a mongoDB query which retrieves every user that has logged in today i.e., whose Last_Login has today's date.

我希望我的输出为:

{
    "_id" : ObjectId("576155a6cd87b68f7e6e42c9"),
    "First_Name" : "ok",
    "Last_Name" : "jaao",
    "Email" : "[e-mail]",
    "Sessions" : [
        {
            "Last_Login" : "Wed, Jun 15, 2016 6:48 PM",
            "ID" : "111",
            "Class" : "fb"
        }
    ],
    "Count" : 2
},
{
    "_id" : ObjectId("576155ccf6d8979e7e77df27"),
    "First_Name" : "abc",
    "Last_Name" : "xyz",
    "Email" : "[e-mail]",
    "Sessions" : [
        {
            "Last_Login" : "Wed, Jun 15, 2016 6:49 PM",
            "Class" : "fb",
            "ID" : "123"
        }
    ],
    "Count" : 1
}

推荐答案

您将需要聚合.

db.users.aggregate([
    {
        $unwind: "$Sessions"
    },
    {
        $match: {
            "Sessions.Last_Login": {
                $gte: ISODate("2016-06-16T00:00:00.0Z"),
                $lt: ISODate("2016-06-17T00:00:00.0Z")
            }
        }
    },
    {
        $group: {
            _id: {
                _id: "$_id",
                First_Name: "$First_Name",
                Last_Name: "$Last_Name"
            },
            Sessions: {
                $push: "$Sessions"
            }
        }
    },
    {
        $project: {
            _id: "$_id._id",
            First_Name: "$_id.First_Name",
            Last_Name: "$_id.Last_Name",
            Sessions: "$Sessions"
        }
    }
])

因此查询将执行以下步骤:

So the query will do those steps:

1. $unwind所有Sessions元素
日期范围内的
2. $match个文档
3. $group和_id，First_Name，Last_Name
一起构成文档
4. $project文档看起来像原始格式

1. $unwind all Sessions elements
2. $match documents inside the date range
3. $group together documents by _id, First_Name, Last_Name
4. $project documents to look like the original format

我省略了一些字段，但是您可以在$group和$project步骤中轻松添加它.当然，您需要更改日期范围.

I omitted some fields, but you can easily add it in $group and $project steps. And of course you'll need to change the date range.

我担心此查询在一个大集合中的性能.如果您使用我给的第一个查询并过滤代码中想要的会话，也许会更好.

I'm concerned about the performance of this query in a big collection. Maybe is better if you use the first query I gave and filter the sessions you want in your code.

正如@chridam所说，仅当将Last_Login更改为ISODate()时，此查询才有效.

As @chridam said, this query will work only if you change Last_Login to ISODate(), what is recommended.

更新查询以使用aggregate并匹配日期范围内仅获取Sessions的请求.

Updating the query to use aggregate and match the request of only fetch Sessions inside the date range.

这是旧版本:

db.users.filter({
    'Sessions': {
        '$elemMatch': {
            'Last_Login': {
                '$gte': ISODate("2016-06-16T00:00:00.0Z"),
                '$lt': ISODate("2016-06-17T00:00:00.0Z")
            }
        }
    }
})

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