如何從Mongoose數組中的每個對象中只選擇一個字段？

Question

我有模式：

{ 
    name: String, 
    surname: String, 
    note: String, 
    someField: String, 
    secondSomeField: String 
    blackList: [{ 
    userId: mongoose.Types.ObjectId, 
    reason: String 
    }] 
} 

我需要的所有字段選擇文件，但在黑名單領域，我需要只選擇用戶ID。示例我想要的：

{ 
     name: "John", 
     surname: "Doe", 
     note: "bwrewer", 
     someField: "saddsa", 
     secondSomeField: "sadsd", 
     blackList: [58176fb7ff8d6518baf0c931, 58176fb7ff8d6518baf0c932, 58176fb7ff8d6518baf0c933, 58176fb7ff8d6518baf0c934] 
} 

我該如何做到這一點？當我做

Schema.find({_id: myCustomUserId}, function(err, user) { 
     //{blackList: [{userId: value}, {userId: value}]} 
     //but i need 
     //{blackList: [value, value, value]} 
}) 

Answer 1

Schema.aggregate({$match: {_id: mongoose.Types.ObjectId(myCustomId)} }, {$project: {blackList: "$blackList.userId", name: true, surname: true, someField: true}}).exec(fn) 

Answer 2

如果你想用默認隱藏字段：

{ 
    name: String, 
    surname: String, 
    note: String, 
    someField: String, 
    secondSomeField: String 
    blackList: [{ 
    userId: mongoose.Types.ObjectId, 
    reason: {type:String, select:false} 
    }] 
} 

如果你只是想在該查詢排除：

Schema 
    .find({_id: myCustomUserId}) 
    .select("-blackList.reason") 
    .exec(function (err, user) { 
     //your filtered results 
}) 

未經測試，但他們應該工作都。