一种可能的方法是使用新聚合框架中的 $group 函数(在 2.1.0 版中可用):
http ://www.mongodb.org/display/DOCS/Aggregation+Framework+-+%24group
> db.images.aggregate({$group:{"_id":"$group", "info":{$push:{"width":"$width", "id":"$_id"}}}})
{
"result" : [
{
"_id" : "2",
"info" : [
{
"width" : "500",
"id" : ObjectId("4f871d51daf6faf42e000001")
},
{
"width" : "150",
"id" : ObjectId("4f871d52daf6faf42e000004")
},
{
"width" : "100",
"id" : ObjectId("4f871d53daf6faf42e000007")
},
{
"width" : "50",
"id" : ObjectId("4f871d53daf6faf42e00000a")
}
]
},
{
"_id" : "1",
"info" : [
{
"width" : "500",
"id" : ObjectId("4f871d4adaf6fa492f000001")
},
{
"width" : "150",
"id" : ObjectId("4f871d4adaf6fa492f000004")
},
{
"width" : "100",
"id" : ObjectId("4f871d4bdaf6fa492f000007")
},
{
"width" : "50",
"id" : ObjectId("4f871d4bdaf6fa492f00000a")
}
]
}
],
"ok" : 1
}
>
这是您希望达到的结果吗?
使用 Map Reduce 操作也可以实现类似的结果:
var map = function(){
var info = [{"width":this.width, "id":this._id}];
emit(this.group, {"info":info});
}
var reduce = function(key, values){
var info = [];
print(key);
print(values.length);
for(var v in values){
print(values[v]);
for(var i in values[v].info){
if(info.indexOf(values[v].info[i]) == -1){
info.push(values[v].info[i]);
};
};
};
return {"info":info};
}
> db.images.mapReduce(map, reduce, {out:{inline:1}})
{
"results" : [
{
"_id" : "1",
"value" : {
"info" : [
{
"width" : "500",
"id" : ObjectId("4f871d4adaf6fa492f000001")
},
{
"width" : "150",
"id" : ObjectId("4f871d4adaf6fa492f000004")
},
{
"width" : "100",
"id" : ObjectId("4f871d4bdaf6fa492f000007")
},
{
"width" : "50",
"id" : ObjectId("4f871d4bdaf6fa492f00000a")
}
]
}
},
{
"_id" : "2",
"value" : {
"info" : [
{
"width" : "500",
"id" : ObjectId("4f871d51daf6faf42e000001")
},
{
"width" : "150",
"id" : ObjectId("4f871d52daf6faf42e000004")
},
{
"width" : "100",
"id" : ObjectId("4f871d53daf6faf42e000007")
},
{
"width" : "50",
"id" : ObjectId("4f871d53daf6faf42e00000a")
}
]
}
}
],
"timeMillis" : 1,
"counts" : {
"input" : 8,
"emit" : 8,
"reduce" : 2,
"output" : 2
},
"ok" : 1,
}
>
希望这将帮助您实现您想要的结果!