mongodb--group聚合运算

1. mongodb本质就是要做一个高性能，能简单则简单，不要把mongodb中的运算做的太复杂
2. count 最简单的一个聚合方法
3. distinct 选择结果中剔除重复的一个键值， 跟sql语句的效果是一样的

   db.person.distinct("age")

1. group 【能不用则不用】 sql的group by
   
   有点像 “AOP“ 编程

db.person.group({
"key":{age:1},
initial:{count:0},
reduce:function(current,aggr){aggr.count++}
})

db.person.group({
... "key":{age:1},
... initial:{count:0},
... reduce:function(current,aggr){aggr.count++}
... })
[
	{
		"age" : 26,
		"count" : 2
	},
	{
		"age" : 29,
		"count" : 2
	},
	{
		"age" : 31,
		"count" : 2
	}
]
{age:26,count:2}
{age:29,count:2}
{age:31,count:2}

再复杂一点：

cond：我们踢掉小于28的数据，那么这时候就只有两个结果了。。。

db.person.group({
"key":{age:1},
initial:{count:0},
reduce:function(current,aggr){aggr.count++},
cond:{"$where":function(){return this.age>28}}
})

db.person.group({
... "key":{age:1},
... initial:{count:0},
... reduce:function(current,aggr){aggr.count++},
... cond:{"$where":function(){return this.age>28}}
... })
[ { "age" : 29, "count" : 2 }, { "age" : 31, "count" : 2 } ]

再复杂一点：

finalizer:

db.person.group({
"key":{age:1},
initial:{count:0},
reduce:function(current,aggr){aggr.count++},
cond:{"$where":function(){return this.age>28}},
finalize:function(current){ return {"age":current.age,"count":current.count,"xxxx":20}}
})

db.person.group({
"key":{age:1},
initial:{count:0},
reduce:function(current,aggr){aggr.count++},
cond:{"$where":function(){return this.age>28}},
finalize:function(current){ current.xxxx=22;  return current;}
})

[ { "age" : 29, "count" : 2,"xxxx":20 }, { "age" : 31, "count" : 2,"xxxx":20  } ]