Question

我有架构：

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

我需要选择包含所有字段的文档，但在blackList字段中我需要只选择userId。示例我想要的：

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

我怎么能这样做？当我做的时候

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

Answer 1

如果您想默认隐藏字段：

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

如果您只想在该查询中排除：

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

未经测试，但他们应该兼顾。

Answer 2

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

如何使用Mongoose从数组中的每个对象中只选择一个字段？

2 个答案: