动态字段数组到单个对象mongo db

Question

我的蒙哥集合如下所示，

{
   "_id" : ObjectId("62aeb8301ed12a14a8873df1"),
   "Fields" : [ 
    {
        "FieldId" : "e8efd0b0-9d10-4584-bb11-5b24f189c03b",
        "Value" : [ 
            "test_123"
        ]
    }, 
    {
        "FieldId" : "fa6745c2-b259-4a3b-8c6f-19eb78fbbbf5",
        "Value" : [ 
            "123"
        ]
    }, 
    {
        "FieldId" : "2a1be5d0-8fb6-4b06-a253-55337bfe4bcd",
        "Value" : []
    }, 
    {
        "FieldId" : "eed12747-0923-4290-b09c-5a05107f5609",
        "Value" : [ 
            "234234234"
        ]
    },        
    {
        "FieldId" : "fe41d8fb-fa18-4fe5-b047-854403aa4d84",
        "Value" : [ 
            "Irrelevan"
        ]
    }, 
    {
        "FieldId" : "93e46476-bf2e-44eb-ac73-134403220e9e",
        "Value" : [ 
            "test"
        ]
    }, 
    {
        "FieldId" : "db434aca-8df3-4caf-bdd7-3ec23252c2c8",
        "Value" : [ 
            "2019-06-16T18:30:00.000Z"
        ]
    }, 
    {
        "FieldId" : "00df903f-5d59-41c1-a3df-60eeafb77d10",
        "Value" : [ 
            "tewt"
        ]
    }, 
    {
        "FieldId" : "e97d0386-cd42-6277-1207-e674c3268cec",
        "Value" : [ 
            "1"
        ]
    },
    {
        "FieldId" : "35e55d27-7d2c-467d-8a88-09ad6c9f5631",
        "Value" : [ 
            "10"
        ]
    }
  ]
 }

这是所有动态表单字段。 因此，我想查询并获得结果喜欢以下对象，

{
    "_id" : ObjectId("62aeb8301ed12a14a8873df1"),
    "e8efd0b0-9d10-4584-bb11-5b24f189c03b": ["test_123"],
    "fa6745c2-b259-4a3b-8c6f-19eb78fbbbf5": ["123"],
    "2a1be5d0-8fb6-4b06-a253-55337bfe4bcd": [],
    "eed12747-0923-4290-b09c-5a05107f5609": ["234234234"],
    "fe41d8fb-fa18-4fe5-b047-854403aa4d84": ["Irrelevan"],
    "93e46476-bf2e-44eb-ac73-134403220e9e":["test"],
    "db434aca-8df3-4caf-bdd7-3ec23252c2c8":["2019-06-16T18:30:00.000Z"],
    "00df903f-5d59-41c1-a3df-60eeafb77d10":["1"]
 }

我希望最终输出像Fields fields的这种组合。FieldId应该是键，字段。值得在这里值得。

请尝试帮助我形成上面的对象。

提前致谢！

Answer 1

您可以使用 $ arraytoObject ，然后使用使用，然后使用作为新的root $ reploceroot ：

db.collection.aggregate([
  {
    $match: {
      // your query here
    }
  },
  {
    $project: {
      newRoot: {
        "$arrayToObject": {
          $map: {
            input: "$Fields",
            in: {
              k: "$this.FieldId",
              v: "$this.Value"
            }
          }
        }
      }
    }
  },
  {
    "$replaceRoot": {
      "newRoot": {
        "$mergeObjects": [
          "$newRoot",
          {
            _id: "$_id"
          }
        ]
      }
    }
  }
])

mongo Playground

Answer 2

我尝试一下并像您想要

 db.collection.aggregate([{
    $replaceWith: {
      $mergeObjects: [
        {
          _id: "$_id"
        },
        {
          $arrayToObject: { $zip: {inputs: ["$Fields.FieldId","$Fields.Value"]}}
        }
      ]
    }
    }])

Playground