如何对聚合隐藏 _id?
要对聚合隐藏 _id,请使用下面的语法:
db.yourCollectionName.aggregate(
{$project : {
_id : 0 ,
yourIncludeFieldName:1,
yourIncludeFieldName:1
}}
).pretty();为了解上述语法,让我们创建一个包含文档的集合。用于创建包含文档的集合的查询如下:
> db.hideidDemo.insertOne({"UserName":"Larry","UserAge":23,"UserCountryName":"US"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b02336de59bd9de06392")
}
> db.hideidDemo.insertOne({"UserName":"Chris","UserAge":21,"UserCountryName":"AUS"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b03036de59bd9de06393")
}
> db.hideidDemo.insertOne({"UserName":"Robert","UserAge":26,"UserCountryName":"UK"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b04036de59bd9de06394")
}借助 find() 方法显示某个集合中的所有文档。查询如下:
> db.hideidDemo.find().pretty();
以下是输出结果:
{
"_id" : ObjectId("5c92b02336de59bd9de06392"),
"UserName" : "Larry",
"UserAge" : 23,
"UserCountryName" : "US"
}
{
"_id" : ObjectId("5c92b03036de59bd9de06393"),
"UserName" : "Chris",
"UserAge" : 21,
"UserCountryName" : "AUS"
}
{
"_id" : ObjectId("5c92b04036de59bd9de06394"),
"UserName" : "Robert",
"UserAge" : 26,
"UserCountryName" : "UK"
}以下是用于对聚合隐藏 _id 的查询:
> db.hideidDemo.aggregate(
... {$project : {
... _id : 0 ,
... UserName:1,
... UserCountryName:1
... }}
... ).pretty();以下是输出结果:
{ "UserName" : "Larry", "UserCountryName" : "US" }
{ "UserName" : "Chris", "UserCountryName" : "AUS" }
{ "UserName" : "Robert", "UserCountryName" : "UK" }
广告
数据结构
网络
RDBMS
操作系统
Java
iOS
HTML
CSS
Android
Python
C 编程语言
C++
C#
MongoDB
MySQL
Javascript
PHP