返回mongodb聚合请求数组的最大值

Question

你好，我有这个json

db.people.insert({
    "ad" : "noc2",
    "createdDate" : ISODate(),
    "list" : [
            {
                    "id" : "p45",
                    "date" : ISODate("2014-01-01T12:18:30.568Z"),
                    "value3" : 21,
                    "value1" : 77,
                    "value2" : 489
            },
            {
                    "id" : "p6",
                    "date" : ISODate("2013-07-18T12:18:30.568Z"),
                    "value3" : 20,
                    "value1" : 20,
                    "value2" : 7
            },
            {
                   "id" : "4578",
                    "date" : ISODate("2013-07-18T12:18:30.568Z"),
                    "value3" : 21,
                    "value1" : 300,
                    "value2" : -319
            }
   ]

})

我想得到我的数组的最大值。例如，我想要一个如下的json：

"result“：{ "_id”："p45"，"value“：587 }

我不工作的聚合请求是：db.test1.aggregate({$match: "ad":"noc2"},{$unwind: '$list'},{$group: {_id: '$ad', list: {recommand: {$max: '$list'}}}});

Answer 1

答案包含查询语法中的几个问题，但大多数情况下都是有效的。下面你可以找到固定的查询：

db.people.aggregate(
  {$match:{ad:"noc2"}},
  {$unwind:"$list"},
  {$project:{_id:0, _id:"$list.id", "value":{$add:["$list.value1","$list.value2","$list.value3"]}}},
  {$sort:{value:-1}},
  {$limit:1}
);

Answer 2

试试这个：

db.people.aggregate([
  {$match:{ad:"noc2"}},
  {$unwind:"$list"},
  {$project:{_id:0, _id":"$list.id", "value":{$add:["$list.value1","$list.value2","$list.value3"]}}},
  {$sort:{value:-1},
  {$limit:1}
])

输出：

{ "result" : [ { "_id" : "p45", "value" : 587 } ], "ok" : 1 }