在mongodb的一个聚合中组合多个组

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问题描述

如果我有这样的收藏:

{
    "store" : "XYZ",
    "total" : 100
},
{
    "store" : "XYZ",
    "total" : 200
},
{
    "store" : "ABC",
    "total" : 300
},
{
    "store" : "ABC",
    "total" : 400
}

我可以通过聚合得到集合中订单的$sum:

I can get the $sum of orders in the collection by aggregation:

db.invoices.aggregate([{$group: { _id: null, total: { $sum: "$total"}}}])

{
    "result": [{
            "_id": null,
            "total": 1000
        }
    ],
    "ok": 1
}

我可以得到按商店分组的订单的$sum:

And I can get the $sum of orders grouped by store:

db.invoices.aggregate([{$group: { _id: "$store", total: { $sum: "$total"}}}])

{
    "result": [{
            "_id": "ABC",
            "total": 700
        }, {
            "_id": "XYZ",
            "total": 300
        }
    ],
    "ok": 1
}

但是我怎样才能在一个查询中做到这一点呢?

But how can I do this in one query?

推荐答案

你可以聚合如下:

$group由store字段，计算出subtotal.

$project 字段 doc 以保持 subtotal 组在下一个组.

$project a field doc to keep the subtotal group in tact, during the nextgroup.

$group by null 并累计净总数.

$group by null and accumulate the net total.

代码:

db.invoices.aggregate([{
            $group: {
                "_id": "$store",
                "subtotal": {
                    $sum: "$total"
                }
            }
        }, {
            $project: {
                "doc": {
                    "_id": "$_id",
                    "total": "$subtotal"
                }
            }
        }, {
            $group: {
                "_id": null,
                "total": {
                    $sum: "$doc.total"
                },
                "result": {
                    $push: "$doc"
                }
            }
        }, {
            $project: {
                "result": 1,
                "_id": 0,
                "total": 1
            }
        }
    ])

输出:

{
    "total": 1000,
    "result": [{
            "_id": "ABC",
            "total": 700
        }, {
            "_id": "XYZ",
            "total": 300
        }
    ]
}

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SUBTOTAL