如何将 MongoDB 集合转换为以元素值为键的对象？

Question

考虑 MongoDB 的 zip codes aggregation 示例 data set。集合中的每个文档如下所示：

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

如何将集合转换为一个对象，其中每个键都是字段的值，每个值都是集合中的对象？

例如，给定两个文档

{ "_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" }

如何将它们转换为单个文档

{
  "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" }
}

?

我正在尝试 MongoDB shell 命令db.zips.aggregate({$project: { "$_id":"$$CURRENT"}}).pretty()，但收到错误消息$expressions are not allowed at the top-level of $project。

我也尝试使用带有命令 db.zips.mapReduce(function(){emit(this._id, this)},function(k,v){}, {out:"stuffs"}) 的 mapReduce，但毫不奇怪，它只会产生（使用 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

您可以使用游标方法 forEach() 来迭代 find() 游标并访问可以添加到对象变量的文档，如以下是两个文档的示例：

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

输出

{
    "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"
    }
}