Question

我正在尝试根据他们的itemName对集合进行分组

这是一个示例json

{
    "_id" : ObjectId("5d7e3f1b2344e220289d2631"),
    "itemName" : "Biscuit",
    "itemSellingPrice" : "10",
    "itemPurchasePrice" : "20",
    "itemAveragePurchasePrice" : "20",
    "itemBaseUnit" : "pcs",
    "itemCategory" : "Bakery",
    "itemReorderPoint" : "10",
    "itemTotalQuantity" : "20",
    "itemSumQuantity" : "20",
    "itemLocation" : "1",
    "itemSubLocation" : "Severino Reyes",
    "itemDateCreated" : ISODate("2019-09-15T13:39:39.650Z"),
    "itemID" : 232,
    "__v" : 0
},
{
    "_id" : ObjectId("5d7e3f202344e220289d2638"),
    "itemName" : "Biscuit",
    "itemSellingPrice" : "10",
    "itemPurchasePrice" : "20",
    "itemAveragePurchasePrice" : "20",
    "itemBaseUnit" : "pcs",
    "itemCategory" : "Bakery",
    "itemReorderPoint" : "10",
    "itemTotalQuantity" : "20",
    "itemSumQuantity" : "100",
    "itemLocation" : "1",
    "itemSubLocation" : "Tomas Mapua",
    "itemDateCreated" : ISODate("2019-09-15T13:39:39.650Z"),
    "itemID" : 233,
    "__v" : 0
}

截至目前，我有8条记录具有相同的“ itemName”：“饼干”

到目前为止，我的查询是这样的

db.getCollection('itemmodels').aggregate( [
   { "$match": { "itemLocation" : "1"} },
    {"$group": 
        {
            "_id": "$itemName",
            "totalQuantity": {"$sum": "$itemSumQuantity"}
        }}
] );

我正在尝试获取这8条记录的$ itemSumQuantity总和。我也试图按原样显示其他数据。

这是我想要的示例输出

{
    "itemName" : "Biscuit",
    "itemSellingPrice" : "10",
    "itemPurchasePrice" : "20",
    "itemAveragePurchasePrice" : "20",
    "itemBaseUnit" : "pcs",
    "itemCategory" : "Bakery",
    "itemReorderPoint" : "10",
    "itemTotalQuantity" : "20",
    "itemSumQuantity" : "120",
    "itemLocation" : "1",
    "numOfDuplicates": 8
    }

因为第一个json数量是20，第二个json数量是100，所以它变成120

知道如何执行此操作吗？我找不到任何group accumulator才能得到想要的结果。

注意：总和对于数量都不起作用

如何在MongoDB中对数据进行分组？

0 个答案: