如何从聚合中隐藏_id？
要从聚合中隐藏_id，请使用以下语法-
db.yourCollectionName.aggregate(
{$project : {
_id : 0 ,
yourIncludeFieldName:1,
yourIncludeFieldName:1
}}
).pretty();为了理解上述语法，让我们用文档创建一个集合。使用文档创建集合的查询如下-
> db.hideidDemo.insertOne({"UserName":"Larry","UserAge":23,"UserCountryName":"US"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b02336de59bd9de06392")
}
> db.hideidDemo.insertOne({"UserName":"Chris","UserAge":21,"UserCountryName":"AUS"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b03036de59bd9de06393")
}
> db.hideidDemo.insertOne({"UserName":"Robert","UserAge":26,"UserCountryName":"UK"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b04036de59bd9de06394")
}在find()方法的帮助下显示集合中的所有文档。查询如下-
> db.hideidDemo.find().pretty();
以下是输出-
{
"_id" : ObjectId("5c92b02336de59bd9de06392"),
"UserName" : "Larry",
"UserAge" : 23,
"UserCountryName" : "US"
}
{
"_id" : ObjectId("5c92b03036de59bd9de06393"),
"UserName" : "Chris",
"UserAge" : 21,
"UserCountryName" : "AUS"
}
{
"_id" : ObjectId("5c92b04036de59bd9de06394"),
"UserName" : "Robert",
"UserAge" : 26,
"UserCountryName" : "UK"
}这是从聚合中隐藏_id的查询-
> db.hideidDemo.aggregate(
... {$project : {
... _id : 0 ,
... UserName:1,
... UserCountryName:1
... }}
... ).pretty();以下是输出-
{ "UserName" : "Larry", "UserCountryName" : "US" }
{ "UserName" : "Chris", "UserCountryName" : "AUS" }
{ "UserName" : "Robert", "UserCountryName" : "UK" }