如何通过MongoDb Shell将字符串字段转换为对象数组中的ObjectId?

时间:2020-08-25 10:46:57

标签: mongodb mongodb-shell

所以我的MongoDb文档看起来像这样:

{
    "_id" : ObjectId("5f1842d238ec58056d94bbb3"),
    "itemsPurchased" : [ 
        {
            "_id" : "5e7365360665d10011756af2",
            "created" : "2020-03-19T12:27:34.108Z",
            "productLine" : "5ce8ec9df5cc72002fee0d4e",
            "price" : 1.9,
            "tax" : 7,
            "ean" : "42272120",
            "expirationDate" : "2020-06-17T12:27:34.103Z",
            "kiosk" : "5c17a3d963ca649138ec522c",
            "loadCell" : "1"
        }, 
        {
            "_id" : "5e7365360665d10011756af3",
            "created" : "2020-03-19T12:27:34.108Z",
            "productLine" : "5ce8ec9df5cc72002fee0d4e",
            "price" : 1.9,
            "tax" : 7,
            "ean" : "42272120",
            "expirationDate" : "2020-06-17T12:27:34.103Z",
            "kiosk" : "5c17a3d963ca649138ec522c",
            "loadCell" : "1"
        }
    ],
    "paymentMethod" : [],
    "type" : "purchase",
    "total" : NumberDecimal("3.8"),
    "session" : ObjectId("5f1842bf1f2028e369d945f0"),
    "orgId" : ObjectId("5cddce9a51cbb2002d636741"),
    "created" : ISODate("2020-07-22T13:44:50.973Z"),
    "updated" : ISODate("2020-07-22T13:44:50.973Z"),
    "__v" : 0
}

问题在于,在itemsPurchased []中(由于逻辑错误),存在id类型的id。例如"_id" : "5e7365360665d10011756af2""productLine" : "5ce8ec9df5cc72002fee0d4e"。如何遍历在itemsPurchased []中具有对象的所有文档,该文档具有字符串而不是ObjectIds,并将其转换为ObjectId类型。我尝试将$ convert和$ toObjectId与$ map一起使用,但似乎无法正确处理。

db.transactions.aggregate([
  { "$match": { "itemsPurchased._id": {$type: "string"}}},
  { "$set": "itemsPurchased1": {
      "$map": {
        "input": "$itemsPurchased",
        "in": {
          "$toObjectId": [
            "$$this",
            {
              "productLine": {
                "$toObjectId": "$$this.productLine"
              }
            }
          ]
        }
      }
    }
  }

无法正确显示地图部分

2 个答案:

答案 0 :(得分:0)

您已正确使用$toObjectId,在$mergeObjects中与$toObjectId合并时,只需更改$$this而不是$map

  {
    "$set": {
      "itemsPurchased1": {
        "$map": {
          "input": "$itemsPurchased",
          "in": {
            "$mergeObjects": ["$$this", { "productLine": { "$toObjectId": "$$this.productLine" } }]
          }
        }
      }
    }
  }

Playground


updateMany

db.transactions.updateMany({}, 
[
  {
    "$set": {
      "itemsPurchased": {
        "$map": {
          "input": "$itemsPurchased",
          "in": {
            "$mergeObjects": ["$$this", { "productLine": { "$toObjectId": "$$this.productLine" } }]
          }
        }
      }
    }
  }
])

答案 1 :(得分:0)

db.getCollection('transactions').update(
{"itemsPurchased.kiosk":{$type: "string"}}, 
[{
    "$set": {
      "itemsPurchased": {
        "$map": {
          "input": "$itemsPurchased",
          "in": {
            "$mergeObjects": [
              "$$this",
              {
                "kiosk": {
                  "$toObjectId": "$$this.kiosk"
                }
              }
            ]
          }
        }
      }
    }
  }], 
{multi:true})

这是更新方式...