仅返回单个属性"_id" [英] return only a single property "_id"

问题描述

我想知道是否有一种方法可以只返回_id，user_id和total而没有items子文档.

I would like to know if there is a way to return only the _id, user_id and total without the items subdocument.

{
    "_id" : 122,
    "user_id" : 123456,
    "total" : 100,
    "items" : [
            {
                    "item_name" : "my_item_one",
                    "price" : 20
            },
            {
                    "item_name" : "my_item_two",
                    "price" : 50
            },
            {
                    "item_name" : "my_item_three",
                    "price" : 30
            }
    ]
}

推荐答案

The second parameter of find lets you select fields. So you can use this (note that the _id field is always selected anyway):

db.mycollection.find({}, {"user_id": 1, "total": 1});

您还可以排除某些字段，因此等效:

You can also exclude certain fields, so this would be equivalent:

db.mycollection.find({}, {"items": 0});

您可以执行以下操作排除_id字段:

You can exclude _id field by doing:

db.mycollection.find({}, {"user_id": 1, "_id": 0});

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