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我在MongoDB中有几个嵌套集合。MongoDB返回平坦结果
当我跑我的服务器上以下查询:
AggregatedData
.aggregateAsync([{
$match: {
"_id.date": {
$gte: dateFrom,
$lte: dateTo
}
}
}, {
$project: {
"date": "$_id.date",
"type": "$_id.type",
"count": "$count.total",
"_id": 0
}
}]);
我结束了这个结果在这里:
[
{
"date": "2016-01-08T00:00:00.000Z",
"type": "V1",
"count": 7359
},
{
"date": "2016-01-08T00:00:00.000Z",
"type": "V2",
"count": 2874
},
{
"date": "2016-01-08T00:00:00.000Z",
"type": "V3",
"count": 512
},
{
"date": "2016-01-07T00:00:00.000Z",
"type": "V1",
"count": 6892
},
{
"date": "2016-01-07T00:00:00.000Z",
"type": "V2",
"count": 3124
},
{
"date": "2016-01-07T00:00:00.000Z",
"type": "V3",
"count": 457
}
]
现在,这就是我想要的:
[
{
"date": "Thu Jan 07 2016 00:00:0 GMT-0800 (PST)",
"types": ["V1", "V2", "V3"],
"values": [7359, 2874, 512]
},
{
"date": "Thu Jan 08 2016 00:00:0 GMT-0800 (PST)",
"types": ["V1", "V2", "V3"],
"values": [6892, 3124, 457]
}
]
我可以改变我的服务器端功能,这是实现:
AggregatedData
.aggregateAsync([{
$match: {
"_id.date": {
$gte: dateFrom,
$lte: dateTo
}
}
}, {
$project: {
"date": "$_id.date",
"type": "$_id.type",
"count": "$count.total",
"_id": 0
}
}])
.then((results) => {
return _.chain(results)
.groupBy('date')
.map(function(value, key) {
return {
date: key,
types: _.pluck(value, 'type'),
values: _.pluck(value, 'count')
}
})
.value();
});
有没有一种方法来达到同样的使用只是MongoDB的聚合框架做服务器端不做处理,让它可以在数据库方面做了什么?
看起来像您需要的项目步骤之后使用[$组(https://docs.mongodb.org/manual/reference/operator/aggregation/group/)运算符 – saljuama
我不知道一吨左右的MongoDB,但似乎为$组和$地图聚合运营商: https://docs.mongodb.org/manual/reference/operator/aggregation/map/ HTTPS:/ /docs.mongodb.org/manual/reference/operator/aggregation/group/ – Vinay