從mongodb的雙嵌套對象數組中獲取多個匹配的元素

Question

主集合文檔包含工作流詳細信息。 每個工作流詳細信息都有一個工作流狀態對象，並且鏈接中有一個對象數組。

當 from.nodeId = "value" 時，我需要從 links 數組中找到所有匹配的對象

{
    "workflowId": "YNmwuXwQKElY",
    "name": "Hello ",
    "workflowState": {
      "searchType": "tasks",
      "links": [
        {
          "id": "67dca090-dd7a-4b86-8522-456ccdb891b2",
          "from": {
            "nodeId": "NecHqBJvxk",
            "portId": "port1"
          },
          "to": {
            "nodeId": "NecHqBJvxk",
            "portId": "port2"
          }
        },
        {
          "id": "9b7a4d78-d4e6-4229-a091-24f87e3e1fa4",
          "from": {
            "nodeId": "PEfgrTpFOB",
            "portId": "port2"
          },
          "to": {
            "nodeId": "nUjhZhIwky",
            "portId": "port1"
          },
          "headerValue": "Identifiers"
        },
        {
          "id": "6e3fb202-7e29-46e0-b940-a4908ae8bb95",
          "from": {
            "nodeId": "NecHqBJvxk",
            "portId": "port2"
          },
          "to": {
            "nodeId": "PEfgrTpFOB",
            "portId": "port1"
          },
          "headerValue": "Conditions",
        }
      ],
      "hovered": {},

    },
    "archive": true
  }

我試過這個查詢

db.collection.find({
  "workflowState.links": {
    $elemMatch: {
      "from.nodeId": "NecHqBJvxk"
    }
  }
},
{
  "workflowState.links.$": 1
})

但它只返回數組中的第一個匹配項。

我期望這種格式的輸出。

{
    "_id": ObjectId("5a934e000102030405000000"),
    "workflowState": {
      "links": [
        {
          "from": {
            "nodeId": "NecHqBJvxk",
            "portId": "port1"
          },
          "id": "67dca090-dd7a-4b86-8522-456ccdb891b2",
          "to": {
            "nodeId": "NecHqBJvxk",
            "portId": "port2"
          }
        },
        {
          "id": "6e3fb202-7e29-46e0-b940-a4908ae8bb95",
          "from": {
            "nodeId": "NecHqBJvxk",
            "portId": "port2"
          },
          "to": {
            "nodeId": "PEfgrTpFOB",
            "portId": "port1"
          },
          "headerValue": "Conditions"
        }
      ]
    }
  }

如何從符合條件的數組中獲取對象的所有元素？

提前致謝

Answer 1

在 $unwind 階段使用聚合函數

   db.collection.aggregate([
        // Match possible documents
        { "$match": {
            "workflowState.links.from.nodeId": "NecHqBJvxk"
        }},

        // Unwind each array
        { "$unwind": "$workflowState.links" },


        // Filter just the matching elements
        { "$match": {
            "workflowState.links.from.nodeId": "NecHqBJvxk"
        }},

        // Group to inner array
        { "$group": {
            "_id": "$_id",
            "links": { "$push":  "$workflowState.links" }
        }},
        { 
            "$project":{
                _id: "$_id",
                "workflowState.links": "$links",
            }
        }
   ]);

Answer 2

要僅獲取數組中的匹配元素，請按如下方式在聚合中使用$filter ：

db.test.aggregate( [
  { 
      $project: {
          "workflowState.links": {
              $filter: {
                   input: "$workflowState.links",
                      as: "link",
                    cond: {
                             $or: [
                                    { $eq: [ "$$link.from.nodeId", "NecHqBJvxk" ] },
                                    { $eq: [ "$$link.to.nodeId", "NecHqBJvxk" ] }
                             ]
                     }
              } 
          }
      }
  }
] )