Node.js获取上周的数据

I have mongoose schema structured somewhat like this:

我有像这样的mongoose架构结构:

{
    "_id": {
        "$oid": "59d00283893f4500127271f6"
    },
    "__v": 0,
    "bills": [
        {
            "timestamp": 1513539218509,
            "table_no": 2,
            "user_id": {
                "$oid": "59cf912782b85b0012860ecc"
            },
            "order_id": {
                "$oid": "5a36c5979a54980014f06787"
            },
            "_id": {
                "$oid": "5a36c6929a54980014f0678b"
            },
            "disc_amount": 320,
            "amount": 320
        },
        {
            "timestamp": 1513539299486,
            "table_no": 2,
            "user_id": {
                "$oid": "59cf912782b85b0012860ecc"
            },
            "order_id": {
                "$oid": "5a36c6c99a54980014f0678c"
            },
            "_id": {
                "$oid": "5a36c6e39a54980014f0678f"
            },
            "disc_amount": 160,
            "amount": 160
        },
        {
            "timestamp": 1513540879109,
            "table_no": 2,
            "user_id": {
                "$oid": "59cf912782b85b0012860ecc"
            },
            "order_id": {
                "$oid": "5a36ccfb9a54980014f06790"
            },
            "_id": {
                "$oid": "5a36cd0f9a54980014f06793"
            },
            "disc_amount": 320,
            "amount": 320
        },
        {
            "timestamp": 1513540986507,
            "table_no": 2,
            "user_id": {
                "$oid": "59cf912782b85b0012860ecc"
            },
            "order_id": {
                "$oid": "5a36cd639a54980014f06794"
            },
            "_id": {
                "$oid": "5a36cd7a9a54980014f06797"
            },
            "disc_amount": 240,
            "amount": 320
        }
    ]
}

From the subschema bills i want to fetch only those bills which are a week old only.

从子模式账单我只想获取那些只有一周的账单。

My first Question is, is it a right way to first find Schema by id, then get its bills and in bills array do processing to fetch only last week data. I guess that won't be efficient because subschema bills can be too large, therefore, it won't be a good idea to fetch whole array. So, what would be the proper way?

我的第一个问题是,它是一种正确的方式,首先通过id找到Schema,然后得到它的账单,并在账单数组中进行处理以仅获取上周的数据。我想这不会有效,因为子模式账单可能太大,因此,获取整个数组不是一个好主意。那么,什么是正确的方法?

And Second question, is there more appropriate way to create schema and do query.(I created subschema bills, so that all bills related to particular id remain at one place, so easy to query. Is this efficient?).

第二个问题,是否有更合适的方法来创建模式和查询。(我创建了子模式账单,以便所有与特定ID相关的账单保留在一个地方,因此很容易查询。这有效吗?)。

Open to all kind of suggestion. Thanks in advance.

接受各种建议。提前致谢。

1 个解决方案

#1

You can create model for your collection in mongoose and use it as following:

您可以在mongoose中为集合创建模型,并按如下方式使用它:

var start = new Date(new Date().getTime() - (7 * 60 * 60 * 24 * 1000));  // get date of last week
  Bill.find({
    'bills.timestamp' : {"$gte": start.getTime()}
  }, {
    // Only include the subdocument(s) that matched the query.
    'bills.$'    : 1
  }, function(err, data){
    console.log(data)
  });

Model:

var mongoose = require('mongoose');

/**
 * Bill Mongo DB model
 * @name billModel
 */
var billModel = function () {

    var billSchema = mongoose.Schema({

    bills:{ type : Array , "default" : [] }

    });


  return mongoose.model('Bill', billSchema);
};

module.exports = new billModel();

For single id, below code can be used:

对于单个id,可以使用以下代码:

Bill.aggregate([
        {
            $match: {
                '_id': new mongoose.Types.ObjectId('5968dcf965854c7817606014')
            }
        },
        {
          $project: {
             bills: {
                $filter: {
                   input: "$bills",
                   as: "bill",
                   cond: { $gte: [ "$$bill.timestamp", start.getTime() ] }
                }
             }
          }
        },
    ], function (err, result) {
        if (err) {
            console.log(err);
        } else {
            res.json(result);
        }
    });

#1