在mongodb中选择"group by"的Max()

Question

请帮我把这个选择句子转换为mongodb:

Select Name, Max(Value) From table1 Group By Name

我读了这篇文章:http://www.mongodb.org/display/DOCS/Aggregation#Aggregation-Group

但仍然不知道如何应用Max()方法而不是SUM()作为该文档.

谢谢.

Answer 1

我创建了Mongo Collection如下.

{ "_id" : ObjectId("4fb36bfd3d1c88bfa15103b1"), "name" : "bob", "value" : 5 }
{ "_id" : ObjectId("4fb36c033d1c88bfa15103b2"), "name" : "bob", "value" : 3 }
{ "_id" : ObjectId("4fb36c063d1c88bfa15103b3"), "name" : "bob", "value" : 7 }
{ "_id" : ObjectId("4fb36c0c3d1c88bfa15103b4"), "name" : "john", "value" : 2 }
{ "_id" : ObjectId("4fb36c103d1c88bfa15103b5"), "name" : "john", "value" : 4 }
{ "_id" : ObjectId("4fb36c143d1c88bfa15103b6"), "name" : "john", "value" : 8 }
{ "_id" : ObjectId("4fb36c163d1c88bfa15103b7"), "name" : "john", "value" : 6 }

然后通过使用以下代码,我按名称和最大值(值)对其进行分组

db.table1.group(
    {key: {name:true},
        reduce: function(obj,prev) { 
            if (prev.maxValue < obj.value) { 
                prev.maxValue = obj.value; 
            }  
        },
    initial: { maxValue: 0 }}
);

结果显示为

[
    {
        "name" : "bob",
        "maxValue" : 7
    },
    {
        "name" : "john",
        "maxValue" : 8
    }
]

聚合框架更简单.您可以使用聚合框架通过以下代码获得相同的结果.

db.table1.aggregate(
    {$group:{_id:"$name", "maxValue": {$max:"$value"}}}
);

Answer 2

使用聚合框架:

db.table1.aggregate({$group:{'_id':'$name', 'max':{$max:'$value'}}}, 
{$sort:{'max':1}}).result

Answer 3

由于 MongoDB 支持 mapreduce，下面的函数应该可以做到。

db.employee.insert({name:"Tejas",Value:2})
db.employee.insert({name:"Tejas",Value:3})
db.employee.insert({name:"Varma",Value:1})
db.employee.insert({name:"Varma",Value:6})

var map=function(){
var key={name:this.name};
var value={value:this.Value};
emit(key,value);
};

var reduce=function(key,values){
var max=-1;
values.forEach(function(value){
            if(max==-1){
        max=value['value'];
    }
    if(max<value['value']){
        max=value['value'];
    }
});
return {max:max};
};

db.employee.mapReduce(map,reduce,{out:{inline:1}});