Mongodb 如何获取按最顶层筛选的文档并对嵌入结果进行排序
假设我有以下文档结构。Mongodb 如何获取按最顶层筛选的文档并对嵌入结果进行排序,mongodb,mongoose,mongodb-query,Mongodb,Mongoose,Mongodb Query,假设我有以下文档结构。 { "_id" : "ehbpidnopgpgcghgakiiiallielefonk", "users" : 11.87, "ratingTopMost" : { "average" : NumberInt("4"), "users" : NumberInt("3174"), }, "reviews" : { "average" : NumberInt("5"), "
{
"_id" : "ehbpidnopgpgcghgakiiiallielefonk",
"users" : 11.87,
"ratingTopMost" : {
"average" : NumberInt("4"),
"users" : NumberInt("3174"),
},
"reviews" : {
"average" : NumberInt("5"),
"count" : NumberInt("51"),
"comments" : [
{
"_id" : ObjectId("5b87b1a07267ad001da7e733"),
"name" : "Batista1395",
"comment" : "Great Stackoverflow",
"datum" : ISODate("2017-01-16T00:00:00.000+01:00"),
"rating" : NumberInt("3")
},
{
"_id" : ObjectId("5b87b1a07267ad001da7e732"),
"name" : "Tim Ace",
"comment" : "I Like it.",
"datum" : ISODate("2016-08-17T00:00:00.000+02:00"),
"bewertung" : NumberInt("5")
},
]
},
}
我需要使用以下条件查询最多20个文档的顶部:
也很好:按Date DESC对嵌入的注释进行排序。预期的输出应该是什么?如代码所示。有大约200条与上述结构相同的记录。预期的输出应该是如上所示的记录,但符合上述条件