MongoDB Aggregation with match显示所有记录

Question

我有如下的json对象。

{
"_id" : ObjectId("123209sfekjern"),
"Name" : "Test1",
"Orders" : [ 
    {
        "Date" : "2020-05-05",
        "Total" : "100.00"
    }, 
    {
        "Date" : "2020-05-10",
        "Total" : "110.00"
    }, 
    {
        "Date" : "2020-05-11",
        "Total" : "100.00"
    }, 
    {
        "Date" : "2020-05-14",
        "Total" : "110.00"
    }, 
    {
        "Date" : "2020-05-20",
        "Total" : "100.00"
    }, 
    {
        "Date" : "2020-05-15",
        "Total" : "100.00"
    }, 
    {
        "Date" : "2020-05-12",
        "Total" : "110.00"
    }, 
    {
        "Date" : "2020-05-18",
        "Total" : "100.00"
    }, 
    {
        "Date" : "2020-05-31",
        "Total" : "110.00"
    }
]

}

我需要所有大于100.00的订单的客户名称，orders.Date和Order.Total

我尝试了以下查询..

db.Customers.aggregate
    (
    [
        {
            $match: {
                $and: [
                    {"Orders.Date":{$gte:"2020-05-15"}},//ISODate('2020-05-15 10:00:00.000Z')
                    { "Orders.Total": { $gte:  "100.00" } },
                    ]
            }
        },
        { $project: { _id:0, Name: 1, "Orders.Total": 1, "Orders.Date": 1} },
    ]
    )

上面的查询返回所有记录。我还是个初学者，正在学习mongodb。任何帮助都将不胜感激。

谢谢。

Answer 1

$match在文档级进行过滤，因此如果至少有一个子文档与您的条件匹配，则会返回整个文档。要过滤嵌套数组，您需要$filter

db.Customers.aggregate([
    {
        $project: {
            _id: 0,
            Name: 1,
            Orders: {
                $filter: {
                    input: "$Orders",
                    cond: {
                        $and: [
                            { $gte: [ "$$this.Date", "2020-05-15" ] },
                            { $gte: [ "$$this.Total", "100.00" ] },
                        ]
                    }
                }
            }
        }
    }
])

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