MongoDB -如何从集合中只检索一个特定的文档

Question

我创建了一个名为"Cars“的数据库，其中包含一个名为"Cars_info”的集合。我在其中插入了8份文件如下。

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" }
])

在这里，我只想从集合中检索第三个文档。但没有匹配任何字段值。喜欢,

db.Cars_info.find({"car_id":"c3"}).pretty()

有办法这样做吗？

Answer 1

你需要.skip()和.limit()。

通过开始索引:2来获取文档，并且只有一个文档，这是第三个文档。

Update：感谢并赞扬@Wernfried指出，您需要.sort()来保证获得文档的第n位。对于您的场景，您必须按照car_id进行排序。

排序一致性

MongoDB不按特定顺序将文档存储在集合中。对包含重复值的字段进行排序时，可以按任何顺序返回包含这些值的文档。

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

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