如何在MongoDB中仅返回单个属性“ _id”？
以下是在MongoDB中仅返回单个属性_id的语法
db.yourCollectionName.find({}, {"_id": 1}).pretty();首先让我们创建一个包含文档的集合
> db.singlePropertyIdDemo.insertOne({"_id":101,"UserName":"Larry","UserAge":21});
{ "acknowledged" : true, "insertedId" : 101 }
> db.singlePropertyIdDemo.insertOne({"_id":102,"UserName":"Mike","UserAge":26});
{ "acknowledged" : true, "insertedId" : 102 }
> db.singlePropertyIdDemo.insertOne({"_id":103,"UserName":"Chris","UserAge":24});
{ "acknowledged" : true, "insertedId" : 103 }
> db.singlePropertyIdDemo.insertOne({"_id":104,"UserName":"Robert","UserAge":23});
{ "acknowledged" : true, "insertedId" : 104 }
> db.singlePropertyIdDemo.insertOne({"_id":105,"UserName":"John","UserAge":27});
{ "acknowledged" : true, "insertedId" : 105 }以下是在find()方法的帮助下显示集合中所有文档的查询
> db.singlePropertyIdDemo.find().pretty();
这将产生以下输出
{ "_id" : 101, "UserName" : "Larry", "UserAge" : 21 }
{ "_id" : 102, "UserName" : "Mike", "UserAge" : 26 }
{ "_id" : 103, "UserName" : "Chris", "UserAge" : 24 }
{ "_id" : 104, "UserName" : "Robert", "UserAge" : 23 }
{ "_id" : 105, "UserName" : "John", "UserAge" : 27 }以下是仅返回单个属性_id的查询
> db.singlePropertyIdDemo.find({}, {"_id": 1}).pretty();这将产生以下输出
{ "_id" : 101 }
{ "_id" : 102 }
{ "_id" : 103 }
{ "_id" : 104 }
{ "_id" : 105 }