mongodb - 通过具有不同值的相同键对MongoDB集合进行排序

Question

我正在尝试查询我的 Mongo 数据库以显示某个集合中的所有值，并按某个键的所有值排序。例如，我有以下收藏:

{
  "id":"1235432",
  "name":"John Smith",
  "occupation":"janitor",
  "salary":"30000"
},
{
  "id":"23412312",
  "name":"Mathew Colins",
  "occupation":"janitor"
  "salary":"32000"
},
{
  "id":"7353452",
  "name":"Depp Jefferson",
  "occupation":"janitor"
  "salary":"33000"
},
{
  "id":"342434212",
  "name":"Clara Smith",
  "occupation":"Accountant",
  "salary":"45000"
},
{
  "id":"794563452",
  "name":"Jonathan Drako",
  "occupation":"Accountant",
  "salary":"46000"
},
{
  "id":"8383747",
  "name":"Simon Well",
  "occupation":"Accountant",
  "salary":"41000"
}

并且我试图只显示职业薪水最高的 TOP 2。我的查询目前看起来像这样:

Stats.find({occupation:{$exists:true}}).populate('name').sort({salary:1}).limit(2)

仅返回 1 个结果，而不是每个职业的一个结果。

How can I change my query to display the top 2 of each occupation by salary range?

Answer 1

您可以使用 $aggregate，如下所述。

 db.collectionName.aggregate({$match:{occupation:{$exists:true}}},
    { $sort: {"salary":-1}},
    { $limit: 2},
   { $project: {"name":1,"salary":1,_id:0} })

输出 JSON:

{"name" : "Jonathan Drako",
"salary" : "46000"},
{"name" : "Clara Smith",
"salary" : "45000"}