Question

主要收集文档包含工作流程详细信息。每个工作流程都有详细信息，其中有一个工作流程状态对象，并且链接中包含一组对象。

当from.nodeId =“ value”时，我需要从links数组中找到所有匹配的对象

{
    "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
  }

我已经尝试过该查询

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

但它仅返回数组中的第一个匹配项。

我希望这种格式的输出。

{
    "_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"
        }
      ]
    }
  }

如何从匹配条件的数组中获取对象的所有元素？

预先感谢

Answer 1

在$ unwind阶段使用Aggregate函数

   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

要仅获取数组中的匹配元素，请按如下所示在汇总中使用$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" ] }
                             ]
                     }
              } 
          }
      }
  }
] )

从对象mongodb的双嵌套数组中获取多个匹配的元素

2 个答案: