mongodb - mongo group query how to keep fields -

everybody. in mongo group query, result shows key(s) in arguments. how keep first document in each group mysql query group. example:

------------------------------------------------------------------------- |  name  | age  |  sex  | province |   city   |   area   |   address     | ------------------------------------------------------------------------- | ddl1st | 22   | 纯爷们 |  beijing |  beijing | chaoyang | qingnianlu    | | ddl1st | 24   | 纯爷们 |  beijing |  beijing | xuhui    | zhaojiabanglu | |  24k   | 220  | ...   |  ....    |  ...     | ...      | ...           | -------------------------------------------------------------------------    db.users.group({key: { name: 1},reduce: function ( curr, result ) { result.count ++ },initial: {count : 0 } }) 

result:

[ {     "name" : "ddl1st",     "count" : 1 }, {     "name" : "24k",     "count" : 1 } ] 

how following:

[    {    "name" : "ddl1st",    "age" : 22,    "sex" : "纯爷们",    "province" : "beijing",    "city" : "beijing",    "area" : "chaoyang",    "address" : "qingnianlu",    "count" : 1    },    {    "name" : "24k",    "age" : 220,    "sex" : "...",    "province" : "...",    "city" : "...",    "area" : "...",    "address" : "...",    "count" : 1 } ] 

if want keep information first matching entries each group, can try aggregating like:

db.test.aggregate({   $group: {     _id: '$name',    name : { $first: '$name' }    age : { $first: '$age' },    sex : { $first: '$sex' },    province : { $first: '$province' },    city : { $first: '$city' },    area : { $first: '$area' },    address : { $first: '$address' },    count: { $sum: 1 }   } }