如何从 mongodb 中的集合中仅检索一个特定文档

Question

我创建了一个名为“Cars”的数据库，并在其中创建了一个名为“Cars_info”的集合。我在里面插入了8个文件如下

db.Cars_info.insertMany([
{car_id: "c1", Company: "Toyota", Model: "Aqua", Year: 2020, Price_in_usd: 25000, Category: "High-end", Country: "Japan"},
{car_id: "c2", Company: "Toyota", Model: "Premio", Year: 2019, Price_in_usd: 35000, Category: "High-end", Country: "Japan"},
{car_id: "c3", Company: "Audi", Model: "A6", Year: 2020, Price_in_usd: 55000, Category: "High-end", Country: "Germany"},
{car_id: "c4", Company: "Tata", Model: "Nano", Year: 2015, Price_in_usd: 10000, Category: "Low-end", Country: "India"},
{car_id: "c5", Company: "Volkswagen", Model: "Taos", Year: 2022, Price_in_usd: 35000, Category: "High-end", Country: "Germany"},
{car_id: "c6", Company: "Ford", Model: "Figo", Year: 2019, Price_in_usd: 26000, Category: "High-end", Country: "America"},
{car_id: "c7", Company: "Mahindra", Model: "Thar", Year: 2018, Price_in_usd: 18000, Category: "Low-end", Country: "India"},
{car_id: "c8", Company: "Honda", Model: "Vezel", Year: 2015, Price_in_usd: 33000, Category: "High-end", Country: "Japan"}
])

这里我只想从集合中检索第三个文档。但是没有匹配任何字段值。像， db.Cars_info.find({"car_id":"c3"}).pretty()
有什么办法吗？

Answer 1

你需要.skip() and .limit().

从索引开始取文档：2，只有1个文档，也就是第三个文档。

更新：感谢并感谢@Wernfried 指出，您需要.sort() 来保证获得文档的第n 个。对于您的场景，您必须按 car_id.

排序

Sort Consistency

MongoDB does not store documents in a collection in a particular order. When sorting on a field that contains duplicate values, documents containing those values may be returned in any order.

db.Cars_info.find()
  .sort({ "car_id": 1 })
  .skip(2)
  .limit(1)

如何从 mongodb 中的集合中仅检索一个特定文档

How to retrieve only one specific document from a collection in mongodb

database

mongodb

mongodb-query

mongo-collection