如何將MongoDB集合轉換爲元素值上的對象？

Question

考慮MongoDB的zip codes aggregation示例data set。集合中的每個文檔如下所示：

{ 
    "_id": "10280", 
    "city": "NEW YORK", 
    "state": "NY", 
    "pop": 5574, 
    "loc": [ 
    -74.016323, 
    40.710537 
    ] 
} 

如何轉變集合到一個對象，其中每個鍵是一個字段的值，並且每個值在集合中的對象？

例如，給出兩個文件

{ "_id" : "01001", "city" : "AGAWAM", "loc" : [ -72.622739, 42.070206 ], "pop" : 15338, "state" : "MA" } 
{ "_id" : "01002", "city" : "CUSHMAN", "loc" : [ -72.51564999999999, 42.377017 ], "pop" : 36963, "state" : "MA" } 

如何轉變成一個單一的文件

{ 
    "01001": { "_id" : "01001", "city" : "AGAWAM", "loc" : [ -72.622739, 42.070206 ], "pop" : 15338, "state" : "MA" }, 
    "01002": { "_id" : "01002", "city" : "CUSHMAN", "loc" : [ -72.51564999999999, 42.377017 ], "pop" : 36963, "state" : "MA" } 
} 

？

我在嘗試MongoDB shell命令db.zips.aggregate({$project: { "$_id":"$$CURRENT"}}).pretty()，我收到錯誤消息$expressions are not allowed at the top-level of $project。

我使用的MapReduce與命令db.zips.mapReduce(function(){emit(this._id, this)},function(k,v){}, {out:"stuffs"})也試過但是，這並不奇怪，只是生產（含db.stuffs.find()）

{ "_id" : "01001", "value" : { "_id" : "01001", "city" : "AGAWAM", "loc" : [ -72.622739, 42.070206 ], "pop" : 15338, "state" : "MA" } } 
{ "_id" : "01002", "value" : { "_id" : "01002", "city" : "CUSHMAN", "loc" : [ -72.51565, 42.377017 ], "pop" : 36963, "state" : "MA" } } 

Answer 1

您可以使用光標方法forEach()迭代的find()光標和訪問然後可以將這些文檔添加到對象變量中，如以下示例中的兩個文檔所示：

var obj = {}; 
var iterator = function(doc){ 
    obj[doc._id] = doc; 
} 
db.collection.find({}).forEach(iterator); 
printjson(obj); 

個

輸出

{ 
    "01001" : { 
     "_id" : "01001", 
     "city" : "AGAWAM", 
     "loc" : [ 
      -72.622739, 
      42.070206 
     ], 
     "pop" : 15338, 
     "state" : "MA" 
    }, 
    "01002" : { 
     "_id" : "01002", 
     "city" : "CUSHMAN", 
     "loc" : [ 
      -72.51565, 
      42.377017 
     ], 
     "pop" : 36963, 
     "state" : "MA" 
    } 
}