Question

我有一种情况，我从聚合中得到一个结果，我以这种格式获取数据。

{
    "_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
    "bacId" : "BAC0023444",
    "cardId" : "2",
    "defaultCardOrder" : "2",
    "alias" : "Finance",
    "label" : "Finance",
    "for" : "",
    "cardTooltip" : {
        "enable" : true,
        "text" : ""
    },
    "dataBlocks" : [
        {
            "defaultBlockOrder" : "1",
            "blockId" : "1",
            "data" : "0"
        },
        {
            "defaultBlockOrder" : "2",
            "blockId" : "2",
            "data" : "0"
        },
        {
            "defaultBlockOrder" : "3",
            "blockId" : "3",
            "data" : "0"
        }
    ],
    "templateBlocks" : [
        {
            "blockId" : "1",
            "label" : "Gross Profit",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "blockId" : "2",
            "label" : "Profit Forecast",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "blockId" : "3",
            "label" : "Resource Billing",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        }
    ]
},
{
    "_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
    "bacId" : "BAC0023444",
    "cardId" : "3",
    "defaultCardOrder" : "3",
    "alias" : "Staffing",
    "label" : "Staffing",
    "for" : "",
    "cardTooltip" : {
        "enable" : true,
        "text" : ""
    },
    "dataBlocks" : [
        {
            "defaultBlockOrder" : "1",
            "blockId" : "1",
            "data" : "1212"
        },
        {
            "defaultBlockOrder" : "2",
            "blockId" : "2",
            "data" : "1120"
        },
        {
            "defaultBlockOrder" : "3",
            "blockId" : "3",
            "data" : "1200"
        }
    ],
    "templateBlocks" : [
        {
            "blockId" : "1",
            "label" : "Staffing Planner",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "blockId" : "2",
            "label" : "Baseline",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "blockId" : "3",
            "label" : "Projected",
            "quarter" : "",
            "data" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        }
    ]
}

现在我想比较每一行的两个对象数组，在这种情况下，它的“dataBlocks”和“templateBlocks”基于“blockId”，我希望得到以下格式的结果。

{
    "_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
    "bacId" : "BAC0023444",
    "cardId" : "2",
    "defaultCardOrder" : "2",
    "alias" : "Finance",
    "label" : "Finance",
    "for" : "",
    "cardTooltip" : {
        "enable" : true,
        "text" : ""
    },
    "blocks" : [
        {
            "defaultBlockOrder" : "1",
            "blockId" : "1",
            "data" : "0",
            "label" : "Gross Profit",
            "quarter" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "defaultBlockOrder" : "2",
            "blockId" : "2",
            "data" : "0",
            "label" : "Profit Forecast",
            "quarter" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "defaultBlockOrder" : "3",
            "blockId" : "3",
            "data" : "0",
            "label" : "Resource Billing",
            "quarter" : "",
            "dataType" : {
                "typeId" : "2"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        }
    ]
},
{
    "_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
    "bacId" : "BAC0023444",
    "cardId" : "3",
    "defaultCardOrder" : "3",
    "alias" : "Staffing",
    "label" : "Staffing",
    "for" : "",
    "cardTooltip" : {
        "enable" : true,
        "text" : ""
    },
    "dataBlocks" : [
        {
            "defaultBlockOrder" : "1",
            "blockId" : "1",
            "data" : "1212",
            "label" : "Staffing Planner",
            "quarter" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "defaultBlockOrder" : "2",
            "blockId" : "2",
            "data" : "1120",
            "label" : "Baseline",
            "quarter" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        },
        {
            "defaultBlockOrder" : "3",
            "blockId" : "3",
            "data" : "1200",
            "label" : "Projected",
            "quarter" : "",
            "dataType" : {
                "typeId" : "1"
            },
            "tooltip" : {
                "enable" : true,
                "text" : ""
            }
        }
    ]
}

是否可以使用mongodb完成它？我使用3.4并尝试使用聚合实现此目的。

提前致谢。

Answer 1

以下查询完成工作：

db.merge.aggregate([
  // unwind twice
  {$unwind: "$templateBlocks"},
  {$unwind: "$dataBlocks"},
  // get rid of documents where dataBlocks.blockId and 
  // templateBlocks.blockId are not equal
  {$redact: {$cond: [{
                        $eq: [
                               "$dataBlocks.blockId",
                               "$templateBlocks.blockId"
                             ]
                      },
                      "$$KEEP",
                      "$$PRUNE"
                    ]
            }
  },
  // merge dataBlocks and templateBlocks into a single document
  {$project: {
                bacId: 1,
                cardId: 1,
                defaultCardOrder: 1,
                alias: 1,
                label: 1,
                for: 1,
                cardTooltip: 1,
                dataBlocks: {
                              defaultBlockOrder: "$dataBlocks.defaultBlockOrder",
                              blockId: "$dataBlocks.blockId",
                              data: "$dataBlocks.data",
                              label: "$templateBlocks.label",
                              quarter: "$templateBlocks.quarter",
                              data: "$templateBlocks.data",
                              dataType: "$templateBlocks.dataType",
                              tooltip: "$templateBlocks.tooltip"
                            }
             }
      },
      // group to put correspondent dataBlocks to an array
      {$group: {
              _id: {
                     _id: "$_id",
                     bacId: "$bacId",
                     cardId: "$cardId",
                     defaultCardOrder: "$defaultCardOrder",
                     alias: "$alias",
                     label: "$label",
                     for: "$for",
                     cardTooltip: "$cardTooltip"
                   },
              dataBlocks: {$push: "$dataBlocks" }
           }
  },
  // remove the unnecessary _id object
  {$project: {
               _id: "$_id._id",
               bacId: "$_id.bacId",
               cardId: "$_id.cardId",
               defaultCardOrder: "$_id.defaultCardOrder",
               alias: "$_id.alias",
               label: "$_id.label",
               for: "$_id.for",
               cardTooltip: "$_id.cardTooltip",
               dataBlocks: "$dataBlocks"
             }
  }
])

考虑到性能取决于数据集的大小，因为查询会展开两次，并且可能会产生大量的中间文档。

Answer 2

您可以在3.6中尝试以下聚合。

下面的查询迭代dataBlocks数组并将数据块元素与模板块元素合并。使用$indexofArray查找模板块，$arrayElemAt使用匹配的块id和db.collection_name.aggregate([{"$addFields":{ "blocks":{ "$map":{ "input":"$dataBlocks", "in":{ "$mergeObjects":[ "$$this", {"$arrayElemAt":[ "$templateBlocks", {"$indexOfArray":["$templateBlocks","$$this.blockId"]} ] } ] } } } }}])来定位数组索引，以访问找到的索引处的元素。

$mergeObjects

对于3.4，将$arrayToObject替换为$objectToArray，$concatArrays和db.collection_name.aggregate([{"$addFields":{ "blocks":{ "$map":{ "input":"$dataBlocks", "in":{ "$arrayToObject":{ "$concatArrays":[ {"$objectToArray":"$$this"}, {"$objectToArray":{ "$arrayElemAt":[ "$templateBlocks", {"$indexOfArray":["$templateBlocks","$$this.blockId"] } ] }} ] } } } } }}])的组合，以合并两个数组中的每个数组元素。

{"$project":{"templateBlocks":0,"dataBlocks":0}}

您可以使用排除项目作为最后一个阶段从输出中删除数组字段。

data Food = Veg Vegetable | Fr Fruit deriving Show

data Fruit = Apple String Bool
            | Cherry String String
            | Grape String deriving Show

data Vegetable = Carrot String
                | Onion Bool String
                | Tomato String String deriving Show


f :: Food -> String
f (Veg v) = fVeg v
f (Fr f) = fFruit f

fVeg (Carrot s) = s
fVeg (Onion b s) = s
fVeg (Tomato s1 s2) = s1 ++ s2

fFruit (Apple s b) = s
...
...

  f $ Veg $ Onion True "friend"
=> "friend"

如何将来自不同数组的两个匹配对象合并为一个对象？

2 个答案: