为此,将aggregate和$zip一起使用。zip用于转置数组。让我们创建一个包含文档的集合-
> db.demo339.insertOne({Id:101,Score1:["98","56"],Score2:[67,89]}); { "acknowledged" : true, "insertedId" : ObjectId("5e529ee5f8647eb59e5620a2") }
在find()方法的帮助下显示集合中的所有文档-
> db.demo339.find();
这将产生以下输出-
{ "_id" : ObjectId("5e529ee5f8647eb59e5620a2"), "Id" : 101, "Score1" : [ "98", "56" ], "Score2" : [ 67, 89 ] }
以下是使用$zip压缩两个数组并创建一个新的对象数组的查询-
> db.demo339.aggregate([ ... { ... "$project": { ... "AllArrayObject": { ... "$map": { ... "input": { ... "$objectToArray": { ... "$arrayToObject": { ... "$zip": { ... "inputs": [ ... "$Score1", ... "$Score2" ... ] ... } ... } ... } ... }, . ... "as": "el", ... "in": { ... "Score1": "$$el.k", ... "Score2": "$$el.v" ... } ... } ... } ... } ... } ... ])
这将产生以下输出-
{ "_id" : ObjectId("5e529ee5f8647eb59e5620a2"), "AllArrayObject" : [ { "Score1" : "98", "Score2" : 67 }, { "Score1" : "56", "Score2" : 89 } ] }