Question

鉴于一系列“购买”，我如何才能获得每个人的最近购买？

每个人在first_name和last_name上都是唯一的。

示例收集：

[
  {
    first_name: "Don",
    last_name: "Foobar",
    date: 11111111, // The unix timestamp of purchase
    purchase: {...}
  },
  {
    first_name: "Don",
    last_name: "Foobar",
    date: 22222222,
    purchase: {...}
  },
  {
    first_name: "James",
    last_name: "McManason",
    date: 12341234,
    purchase: {...}
  }
  ...
]

我尝试了什么：

下面的代码可以工作（超级次优），给出一组人名来迭代：

collection
  .find({ first_name: "Tom", last_name: "Brady" })
  .sort({ date: -1 })
  .limit(1)

collection
  .find({ first_name: "James", last_name: "Foo" })
  .sort({ date: -1 })
  .limit(1)

collection
  .find({ first_name: "Marcia", last_name: "Bar" })
  .sort({ date: -1 })
  .limit(1)

Answer 1

那么您需要更通用的解决方案吗？如果是这样，那就试试吧：

db.collection.aggregate([
{ $sort: { date: -1}}, // sort by date descending
{
  $group: {
    _id: { firstName: "$first_name", lastName: "$last_name"}, // group by
                                                             // first and last name
    purchase: {$first: "$purchase"} // get the first purchase since the documents are 
                                    // ordered by date and the first is also the latest
  }
}
])

对整个集合进行排序虽然效率不高，但您应该考虑在$match之前添加$sort。

获取MongoDB中多个字段的最新记录

1 个答案: