MongoDb 仅当数组不为空时才在数组中添加字段

Question

现在我有了新情况3.0版。 我有这个假的 json：

[
   {
      "type":"PF",
      "code":12345,
      "Name":"Darth Vader",
      "currency":"BRL",
      "status":"ACTIVE",
      "localization":"NABOO",
      "createDate":1627990848665,
      "olderAdress":[
         {
            "localization":"DEATH STAR",
            "status":"BLOCKED",
            "createDate":1627990848665
         },
         {
            "localization":"TATOOINE",
            "status":"CANCELLED",
            "createDate":1627990555665
         },
         {
            "localization":"ALDERAAN",
            "status":"INACTIVED",
            "createDate":1627990555665
         }
      ]
   },
   {
      "type":"PF",
      "code":12345,
      "Name":"Anakin Skywalker",
      "currency":"BRL",
      "status":"ACTIVE",
      "localization":"NABOO",
      "createDate":1627990848665,
      "olderAdress":null
   }
]

我需要在每个数组元素中添加一个新字段仅当数组不为空时。但我需要通过聚合来做到这一点，因为我在发送给用户之前在 Spring 中使用了这个结果。

我需要这个结果：

[
  {

      "type": "PF",
      "code": 12345,
      "Name": "Darth Vader",
      "currency": "BRL",
      "status": "ACTIVE",
      "localization": "NABOO",
      "createDate": 1627990848665,
      "olderAddress": [
        {
          "localization": "DEATH STAR",
          "status": "BLOCKED",
          "createDate": 1627990848665,
          "isItemOfOlderAddress" : true
        },
        {
          "localization": "TATOOINE",
          "status": "CANCELLED",
          "createDate": 1627990555665,
          "isItemOfOlderAddress" : true
        },
        {
          "localization": "ALDERAAN",
          "status": "INACTIVED",
          "createDate": 1627990555665,
          "isItemOfOlderAddress" : true
        },
      ]
    },
  {
      "type": "PF",
      "code": 12345,
      "Name": "Anakin Skywalker",
      "currency": "BRL",
      "status": "ACTIVE",
      "localization": "NABOO",
      "createDate": 1627990848665,
      "olderAdress": null
    },  
]

所以我只在 olderAddress 不为 null 和 oldAddress 为 null 的地方添加了字段 isItemOfOlderAddress 我只显示默认信息。我该怎么做？

Answer 1

查询

• 如果 oldAdress 是 array（所以也不是 null），将 "isItemOfOlderAddress": true 字段添加到所有成员
• 否则保留旧值（所以也保留null）

Test code here

db.collection.aggregate([
  {
    "$set": {
      "olderAdress": {
        "$cond": [
          {
            "$isArray": [
              "$olderAdress"
            ]
          },
          {
            "$map": {
              "input": "$olderAdress",
              "in": {
                "$mergeObjects": [
                  "$$this",
                  {
                    "isItemOfOlderAddress": true
                  }
                ]
              }
            }
          },
          "$olderAdress"
        ]
      }
    }
  }
])