Mongodb如何在单个查询中找到两个数组中的共同点并按降序排序

Question

这是我保存在数据库中的数据

0  _id:5e4d18bd10e5482eb623c6e4
   name:'John singh',
   cars_owned:[
       {car_id:'1'},
       {car_id:'5'},
       {car_id:'7'},
       {car_id:'8'}
   ],

1  _id:5e4d18bd10e5482eb6g57f5rt
   name:'Josh kumar',
   cars_owned:[
       {car_id:'7'},
       {car_id:'9'},
       {car_id:'1'},
       {car_id:'3'}
   ],

2  _id:5e4d18bd10e5482eb6r67222
   name:'Jesse nigam',
   cars_owned:[
       {car_id:'6'},
       {car_id:'7'},
       {car_id:'9'},
       {car_id:'3'}
   ],

3  _id:5e4d18bd10e5482eb6467ii46
   name:'Jordan khan',
   cars_owned:[
       {car_id:'3'},
       {car_id:'1'},
       {car_id:'4'},
       {car_id:'5'}
   ]

现在我想搜索一个用户，其起始名称为“J”，并且我的 cars_owned 输入将是

  'J',cars_owned['3','7','9','12','10']

输出将是

   1  _id:5e4d18bd10e5482eb6g57f5rt
   name:'Josh kumar',
   cars_owned:[
       {car_id:'7'},
       {car_id:'9'},
       {car_id:'1'},
       {car_id:'3'}
   ],

   2  _id:5e4d18bd10e5482eb6r67222
   name:'Jesse nigam',
   cars_owned:[
       {car_id:'6'},
       {car_id:'7'},
       {car_id:'9'},
       {car_id:'3'}
   ],

   0  _id:5e4d18bd10e5482eb623c6e4
   name:'John singh',
   cars_owned:[
       {car_id:'1'},
       {car_id:'5'},
       {car_id:'7'},
       {car_id:'8'}
   ],

   3  _id:5e4d18bd10e5482eb6467ii46
   name:'Jordan khan',
   cars_owned:[
       {car_id:'3'},
       {car_id:'1'},
       {car_id:'4'},
       {car_id:'5'}
   ]

现在您会注意到结果是所有名称以“J”开头的用户，cars_owned 是在 3,7,9,12,10 但降序排列的用户是大多数cars_owned 都在顶部和其他方面相匹配。我希望根据单个 mongo db 查询中匹配的最大 cars_owned 对结果进行排序。到目前为止，我已经做了一个简单的查找查询。

  User_data.find({name: { $regex: "^" + search_name },
                  cars_owned:{$elemMatch:{car_id:'3',car_id:'7',car_id:'9',car_id:'12',car_id:'10'} 
                 }},function(err,resp){

                      console.log(JSON.stringify(resp,null,4));

          });

但它只返回一个文档我想要所有具有任何给定 car_id 但按最大匹配排序的文档。如果您对这个问题有任何不理解，请随时在评论中提问，但请在单个 mongo db 查询中给出答案，我也可以使用 aggeregation 框架。提前致谢。

Answer 1

您必须为此使用聚合。 首先，您定义列表中有多少用户的汽车是常见的，然后过滤匹配名称和至少一辆汽车常见的结果，最后按常见汽车对结果进行排序。

这里是查询：

db.collection.aggregate([
  {
    $addFields: {
      commonCars: {
        $size: {
          $setIntersection: [
            [
              "3",
              "7",
              "9",
              "12",
              "10"
            ],
            "$cars_owned.car_id"
          ]
        },

      }
    }
  },
  {
    $match: {
      $expr: {
        $and: [
          {
            $eq: [
              {
                $regexMatch: {
                  input: "$name",
                  regex: "^J"
                }
              },
              true
            ]
          },
          {
            $gt: [
              "$commonCars",
              0
            ]
          }
        ]
      }
    }
  },
  {
    $sort: {
      "commonCars": -1
    }
  }
])

And you can test it here

---

编辑

如果您不需要对结果进行排序，您可以在一个匹配阶段实现：

db.collection.aggregate([
  {
    $match: {
      $expr: {
        $and: [
          {
            $eq: [
              {
                $regexMatch: {
                  input: "$name",
                  regex: "^J"
                }
              },
              true
            ]
          },
          {
            $gt: [
              {
                $size: {
                  $setIntersection: [
                    [
                      "3",
                      "7",
                      "9",
                      "12",
                      "10"
                    ],
                    "$cars_owned.car_id"
                  ]
                },

              },
              0
            ]
          }
        ]
      }
    }
  },

])

Test it here