MongoDB聚合操作：如何将字符串数组转为对象数组并拆分assets字段

原数据结构 #

[
  {
    "_id": ObjectId("5c949609ff5e3d6119758730"),
    "assets": [
      "one##two##three##four",
      "one##two##hello##world"
    ]
  }
]

期望转换结果 #

[
  {
    "_id": ObjectId("5c949609ff5e3d6119758730"),
    "assets": [
        {
            "firstkey": "one",
            "secondkey": "two",
            "thirdkey": "three",
            "fourthkey": "four"
        },
        {
            "firstkey": "one",
            "secondkey": "two",
            "thirdkey": "hello",
            "fourthkey": "world"
        }
    ]
  }
]

聚合查询实现方法 #

方法一：直接构造自定义键对象

通过$map遍历assets数组，对每个字符串用$split拆分后，用$arrayElemAt取出对应位置的值并赋值给自定义键：

db.collection.aggregate([
  {
    $addFields: {
      assets: {
        $map: {
          input: "$assets",
          as: "item",
          in: {
            $let: {
              vars: {
                parts: { $split: ["$$item", "##"] }
              },
              in: {
                firstkey: { $arrayElemAt: ["$$parts", 0] },
                secondkey: { $arrayElemAt: ["$$parts", 1] },
                thirdkey: { $arrayElemAt: ["$$parts", 2] },
                fourthkey: { $arrayElemAt: ["$$parts", 3] }
              }
            }
          }
        }
      }
    }
  }
])

方法二：用$zip+$arrayToObject动态生成对象

如果自定义键名数量较多，这种方式更简洁：将键名数组和拆分后的字符串数组合并，再转成对象：

db.collection.aggregate([
  {
    $addFields: {
      assets: {
        $map: {
          input: "$assets",
          as: "item",
          in: {
            $arrayToObject: {
              $zip: {
                inputs: [["firstkey", "secondkey", "thirdkey", "fourthkey"], { $split: ["$$item", "##"] }]
              }
            }
          }
        }
      }
    }
  }
])

说明 #

两种方法都能在聚合查询中直接定义firstkey、secondkey等自定义键名，完成从字符串到目标对象的转换。

内容的提问来源于stack exchange，提问作者xxestter