如何對聚合隱藏 _id?
要對聚合隱藏 _id,請使用下面的語法:
db.yourCollectionName.aggregate(
{$project : {
_id : 0 ,
yourIncludeFieldName:1,
yourIncludeFieldName:1
}}
).pretty();為了解上述語法,讓我們建立一個包含文件的集合。用於建立包含文件的集合的查詢如下:
> db.hideidDemo.insertOne({"UserName":"Larry","UserAge":23,"UserCountryName":"US"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b02336de59bd9de06392")
}
> db.hideidDemo.insertOne({"UserName":"Chris","UserAge":21,"UserCountryName":"AUS"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b03036de59bd9de06393")
}
> db.hideidDemo.insertOne({"UserName":"Robert","UserAge":26,"UserCountryName":"UK"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c92b04036de59bd9de06394")
}藉助 find() 方法顯示某個集合中的所有文件。查詢如下:
> db.hideidDemo.find().pretty();
以下是輸出結果:
{
"_id" : ObjectId("5c92b02336de59bd9de06392"),
"UserName" : "Larry",
"UserAge" : 23,
"UserCountryName" : "US"
}
{
"_id" : ObjectId("5c92b03036de59bd9de06393"),
"UserName" : "Chris",
"UserAge" : 21,
"UserCountryName" : "AUS"
}
{
"_id" : ObjectId("5c92b04036de59bd9de06394"),
"UserName" : "Robert",
"UserAge" : 26,
"UserCountryName" : "UK"
}以下是用於對聚合隱藏 _id 的查詢:
> db.hideidDemo.aggregate(
... {$project : {
... _id : 0 ,
... UserName:1,
... UserCountryName:1
... }}
... ).pretty();以下是輸出結果:
{ "UserName" : "Larry", "UserCountryName" : "US" }
{ "UserName" : "Chris", "UserCountryName" : "AUS" }
{ "UserName" : "Robert", "UserCountryName" : "UK" }
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