Question

这是单个文档：

function toNumber(str) {
  str = String(str).trim();
  return !str ? NaN : Number(str);
}
console.log(!isNaN(toNumber("10000"))); // true
console.log(!isNaN(toNumber("100T0"))); // false
console.log(!isNaN(toNumber("")));      // false

我试图仅选择John's BMW的“ carColor”。像这样：

{
   _id: "...",
   firstName: "john",
   lastName:"Doe",
   cars: [
       {
           "_id": "...",
           "carName": "BMW",
           "carModel": "330",
           "carColor": "silver"
       },
       {
           "_id": "...",
           "carName": "Lexus",
           "carModel": "IS300",
           "carColor": "white"
       },
       {
           "_id": "...",
           "carName": "LADA",
           "carModel": "2106",
           "carColor": "blue"
       }
   ]
}

但是此查询返回完整的对象，如下所示：

db.persons.findOne(
        { "firstName": "John", "cars.carName": "BMW" },
        { "_id": 0, "cars.$.carColor": 1 }
      );

我已经尝试了没有。$的其他查询。符号：

{
    cars: [
      {
         "_id": "...",
         "carName": "BMW",
         "carModel": "330",
         "carColor": "silver"
      }
}

此版本仅返回“ carColor”属性，但不过滤“ carName”。像这样：

db.persons.findOne(
            { "firstName": "John", "cars.carName": "BMW" },
            { "_id": 0, "cars.carColor": 1 }
          );

有什么想法吗？

Answer 1

为什么不起作用？

{“ firstName”：“ John”，“ cars.carName”：“ BMW”}

表示“名称为john，并且在汽车数组中至少有一个条目，其中carName为” BMW”的地方”。但是它返回完整的文档，而不过滤数组。

{“ _id”：0，“ cars.carColor”：1}

不投影_id，而是投影car数组所有条目的carColor。

解决方案

实际上，使用查找和投影方法无法完全实现所需的功能。最好做的是像这样添加$ projection operator：

db.collection.find({
  firstName: "john",
  "cars.carName": "BMW"
},
{
  _id: 0,
      "cars.$": 1
    })

**RESULT**

[
  {
    "cars": [
      {
        "_id": "...",
        "carColor": "silver",
        "carModel": "330",
        "carName": "BMW"
      }
    ]
  }
]

但是这种方法有缺点：

您可以获得整个数组条目，而不仅是想要/需要的颜色
它仅返回第一个匹配项：如果john有2个BMW，则仅返回一个。

更好的解决方案

幸运的是，MongoDB通过聚合框架和$filter运算符提供了另一种实现此目标的方法：

db.collection.aggregate([
  {
    $match: {
      firstName: "john"
    }
  },
  {
    $project: {
      cars: {
        $filter: {
          input: "$cars",
          as: "cars",
          cond: {
            $eq: [
              "$$cars.carName",
              "BMW"
            ]
          }
        }
      }
    }
  },
  {
    $project: {
      _id: 0,
      "colors": "$cars.carColor"
    }
  }
])

You can try it here.

编辑：其他解决方案

您也可以通过展开/分组阶段来尝试：

db.collection.aggregate([
  {
    $match: {
      firstName: "john"
    }
  },
  {
    $unwind: "$cars"
  },
  {
    $match: {
      "cars.carName": "BMW"
    }
  },
  {
    $group: {
      "_id": null,
      colors: {
        $push: "$cars.carColor"
      }
    }
  }
])

Answer 2

db.persons.find({
  firstName: 'john',
  cars: {
    $elemMatch: {
      carName: 'BMW'
    }
  }
},
{
  'cars.$': 1
})

Answer 3

如果您知道数组中的“ BMW”值不超过一个，那么这是一种通过单个$project阶段获得结果的方法：

db.getCollection('collection').aggregate([{
    $match: {
        "firstName": "john"
        /* for performance reasons, you may want to include the following line which, however, is not required */
        /* this makes sense if you have lots of "john"s with different sets of cars in your database */
        , "cars.carName": "BMW" // this will use an index on "cars.carName" if available
    }
}, {
    $project: {
        _id: 0, // do not return the _id field
        color: {
            $reduce: { // transform the filtered input array
                "input": {
                    $filter: { // remove all non-"BMW" cars from the "cars" array
                        input: "$cars",
                        as: "car",
                        cond: { $eq: [ "$$car.carName", "BMW" ] }
                    }
                },
                "initialValue": null,
                "in": "$$this.carColor" // just return the color value, nothing else
            }
        }
    }
}])

如何在MongoDB中使用多个条件查找查询指定字段？

3 个答案: