在 MongoDB 中合并来自 2 个集合的文档并保留字段的属性答案

【问题标题】：Merge documents from 2 collections in MongoDB & preserve property of a field在 MongoDB 中合并来自 2 个集合的文档并保留字段的属性
【发布时间】：2022-01-18 14:06:51
【问题描述】：

我有两个集合，1.temporaryCollection，2.permanentCollection，我想从temporaryCollection 中获取数据并在permanentCollection 中更新。要查看预期结果，请参阅下面的 updatedPermanentCollection。

从临时集合中获取并在永久集合中更新的字段是：

电子邮件地址
电话号码
联系人姓名
联系电话

临时集合中更改的字段供您参考

联系人[0]['emailAddresses']
联系人[0]['ContactName']
联系人[0]["phoneNumbers"]
联系人[0]["ContactNumber"]

UpdatedPermanentCollection 中更新后不应更改的字段是

contacts._id

注意：contacts 是一个对象数组，为简单起见，我只展示了一个对象。

我目前正在使用以下查询，该查询更新了 PermanentCollection，但也覆盖了 contacts._id 字段。我不希望 contacts._id 字段被覆盖。

这是我的 MongoDB 查询

db.temporaryCollection.aggregate([
  {
    $match: {
      userID: ObjectId("61d1efea2c0fab00340f47c8"),
    },
  },
  {
    $merge: {
      into: "permanentCollection",
      on: "userID",
      whenMatched: "merge",
      whenNotMatched: "insert",
    },
  },
]);

1.临时集合

{
  "_id": { "$oid": "61d1f04266289f003452d705" },
  "userID": { "$oid": "61d1efea2c0fab00340f47c8" },
  "contacts": [
    {
      "emailAddresses": [
        { "id": "6884", "label": "email1", "email": "addedemail@gmail.com" }
      ],
      "phoneNumbers": [
        {
          "label": "other",
          "id": "4594",
          "number": "+918984292930"
        },
        {
          "label": "other",
          "id": "4595",
          "number": "+911234567890"
        }
      ],
      "_id": { "$oid": "61d1f04266289f003452d744" },
      "ContactName": "Sample User 1 Name Changed",
      "ContactNumber": "+918984292930",
      "recordID": "833"
    }
  ],
  "userNumber": "+911234567890",
  "__v": 7
}

2。永久收藏

    {
  "_id": { "$oid": "61d1f04266289f003452d701" },
  "userID": { "$oid": "61d1efea2c0fab00340f47c8" },
  "contacts": [
    {
      "emailAddresses": [],
      "phoneNumbers": [
        {
          "label": "other",
          "id": "4594",
          "number": "+918984292929"
        },
        {
          "label": "other",
          "id": "4595",
          "number": "+911234567890"
        }
      ],
      "_id": { "$oid": "61d1f04266289f003452d722" },
      "ContactName": "Sample User 1",
      "ContactNumber": "+918984292929",
      "recordID": "833"
    }
  ],
  "userNumber": "+911234567890",
  "__v": 7
}

3。 updatedPermanentCollection（预期结果）

    {
  "_id": { "$oid": "61d1f04266289f003452d701" },
  "userID": { "$oid": "61d1efea2c0fab00340f47c8" },
  "contacts": [
    {
      "emailAddresses": [
        { "id": "6884", "label": "email1", "email": "addedemail@gmail.com" }
      ],
      "phoneNumbers": [
        {
          "label": "other",
          "id": "4594",
          "number": "+918984292930"
        },
        {
          "label": "other",
          "id": "4595",
          "number": "+911234567890"
        }
      ],
      "_id": { "$oid": "61d1f04266289f003452d722" },
      "ContactName": "Sample User 1 Name Changed",
      "ContactNumber": "+918984292930",
      "recordID": "833"
    }
  ],
  "userNumber": "+911234567890",
  "__v": 7
}

【问题讨论】：

标签： mongodb mongoose mongodb-query

【解决方案1】：

试试这个聚合查询。

db.temporarCollection.aggreagate(
[
  {
    "$lookup": {
      "from": "permanantCollection", 
      "let": {
        "user_id": "$userID"
      }, 
      "pipeline": [
        {
          "$match": {
            "$expr": {
              "$eq": [
                "$$user_id", "$userID"
              ]
            }
          }
        }
      ], 
      "as": "pcontacts"
    }
  }, {
    "$unwind": {
      "path": "$pcontacts", 
      "preserveNullAndEmptyArrays": true
    }
  }, {
    "$project": {
      "contacts": {
        "$map": {
          "input": "$contacts", 
          "as": "contact", 
          "in": {
            "tcontact": "$$contact", 
            "pcontact": {
              "$first": {
                "$filter": {
                  "input": "$pcontacts.contacts", 
                  "as": "pcontact", 
                  "cond": {
                    "$eq": [
                      "$$pcontact.recordID", "$$contact.recordID"
                    ]
                  }
                }
              }
            }
          }
        }
      }, 
      "userNumber": 1, 
      "userID": 1, 
      "_id": 0
    }
  }, {
    "$project": {
      "contacts": {
        "$map": {
          "input": "$contacts", 
          "as": "contact", 
          "in": {
            "emailAddresses": "$$contact.tcontact.emailAddresses", 
            "phoneNumbers": "$$contact.tcontact.phoneNumbers", 
            "ContactName": "$$contact.tcontact.ContactName", 
            "ContactNumber": "$$contact.tcontact.ContactNumber", 
            "recordID": {
              "$let": {
                "vars": {}, 
                "in": {
                  "$cond": {
                    "if": "$$contact.pcontact.recordID", 
                    "then": "$$contact.pcontact.recordID", 
                    "else": "$$contact.tcontact.recordID"
                  }
                }
              }
            }, 
            "_id": {
              "$let": {
                "vars": {}, 
                "in": {
                  "$cond": {
                    "if": "$$contact.pcontact._id", 
                    "then": "$$contact.pcontact._id", 
                    "else": "$$contact.tcontact._id"
                  }
                }
              }
            }
          }
        }
      }, 
      "userNumber": 1, 
      "userID": 1
    }
  }, {
    "$merge": {
      "into": "pc", 
      "on": "userID", 
      "whenMatched": "replace", 
      "whenNotMatched": "insert"
    }
  }
])

这不是一个完全优化的查询，但它可以工作。

【讨论】：

【解决方案2】：

尝试将 $unset 添加到数据库查询中。

db.temporaryCollection.aggregate([
  {
    $unset: "_id"
  },
  {
    $match: {
      userID: ObjectId("61d1efea2c0fab00340f47c8"),
    },
  },
  {
    $merge: {
      into: "permanentCollection",
      on: "userID",
      whenMatched: "merge",
      whenNotMatched: "insert",
    },
  },
]);

【讨论】：