MongoDB Reduce Key with single value

Question

我想计算客户的订单数量，因此，在去年，上个月和上周。我写了一个MapReduce程序：

var mapOrders = function() {
    var v_order = {
        order_date : this.dt_order
        ...
    }; 
     emit(this.clientid, v_order);
};
var reduceOrders = function(p_clientid, p_orders) {
    // Initialization of the output format of the couters
    var r_result = { orders_count : {
        total: {
            1year: 0,
            1month: 0,
            7day: 0
        }
        ...
    }}
    for (var c_order = 0; c_order < p_orders.length; c_order++) {
        // Increment counters
    }
    return (r_result);
};
db.orders.mapReduce(
    mapOrders,
    reduceOrders,
    { 
        out: { merge:  "tmp_orders_indicators" }
    }
)

在我的输出集合中，我有 2 种类型的记录

{
    "_id" : 80320,
    "value" : {
        "order_date" : ISODate("2015-10-30T11:09:51.000Z")
        ...
    }
}
{
    "_id" : 80306,
    "value" : {
        "orders_count" : {
            "total" : {
                "count_1year" : 18,
                "count_1month" : 6,
                "count_7day" : 1
            }
            ...
        }
}

只有 1 个订单的客户端不会通过 reduce 功能。我在MongoDB的双重处理中发现了这个，它解释了这种行为：

MongoDB不会为只有 单个值。

如何使输出集合中只有 1 种类型的记录如下所示？强制所有记录通过reduce函数？

{
    "_id" : 80306,
    "value" : {
        "orders_count" : {
            "total" : {
                "count_1year" : 18,
                "count_1month" : 6,
                "count_7day" : 1
            }
            ...
        }
}

Answer 1

您可以通过聚合无缝实现此目的。请考虑以下管道：

var dateSevenDaysAgo = new Date();
dateSevenDaysAgo.setDate(dateSevenDaysAgo.getDate()-7);
var dateMonthAgo = new Date();
dateMonthAgo.setMonth(dateMonthAgo.getMonth()-1);
var dateYearAgo = new Date();
dateYearAgo.setFullYear(dateYearAgo.getFullYear()-1);
var pipeline = [
    { "$match": { "$dt_order": { "$gte": dateYearAgo } } },
    {
        "$group": {
            "_id": "$id_client",
            "count_1year": {
                "$sum": {
                    "$cond": [ 
                        { "$gte": [ "$dt_order", dateYearAgo ] }, 
                        1, 0 
                    ]
                }
            },
            "count_1month": {
                "$sum": {
                    "$cond": [ 
                        { "$gte": [ "$dt_order", dateMonthAgo ] }, 
                        1, 0 
                    ]
                }
            },
            "count_7day": {
                "$sum": {
                    "$cond": [ 
                        { "$gte": [ "$dt_order", dateSevenDaysAgo ] }, 
                        1, 0 
                    ]
                }
            }
        }
    },
    { "$out": "tmp_indicators" }
];
db.orders.aggregate(pipeline);
db.tmp_indicators.find();

Answer 2

使用 finalize 实用程序找到了解决方案。

var mapOrders = function() {
    var v_order = {
        order_date : this.dt_order
        ...
    }; 
     emit(this.clientid, v_order);
};
var reduceOrders = function(p_clientid, p_orders) {
    // Initialization of the output format of the couters
    var r_result = { orders_count : {
        total: {
            1year: 0,
            1month: 0,
            7day: 0
        }
        ...
    }}
    for (var c_order = 0; c_order < p_orders.length; c_order++) {
        // Increment counters
    }
    return (r_result);
};

var finalizeOrders = function(p_clientid, p_ReducedDrders) {
    if (typeof p_ReducedDrders.orders_count === 'undefined' )
        // Initialization of the output format of the couters
        var r_result = { orders_count : {
            total: {
                1year: 0,
                1month: 0,
                7day: 0
            }
        ...
        }}
        // do the same stuff as the for loop in the reducer
    }
    else {
        r_result = p_ReducedDrders
    }
    return (r_result);
};
db.orders.mapReduce(
    mapOrders,
    reduceOrders,
    { 
        out: { merge:  "tmp_orders_indicators" },
        finalize : finalizeOrders
    }
)