Mongo：获取具有匹配条件的每个组的第一个文档

Question

我有带有日期，txt，发送者和接收者的消息模型。我需要获取两个用户之间的最后一条消息，例如：

{_id: 1, sender:1, receiver:2, txt: “hi 2”, date: “2018/04/01”}
{_id: 2, sender:1, receiver:2, txt: “how are you 2”, date: “2018/04/02”}
{_id: 3, sender:2, receiver:1, txt: “hi 1”, date: “2018/04/03”}
{_id: 4, sender:1, receiver:3, txt: “hi 3”, date: “2018/04/01”}
{_id: 5, sender:1, receiver:3, txt: “how are you 3”, date: “2018/04/02”}

如果我要查找涉及用户1的新邮件，则需要：

{_id: 3, sender:2, receiver:1, txt: “hi 1”, date: “2018/04/03”}
{_id: 5, sender:1, receiver:3, txt: “how are you 3”, date: “2018/04/02”}

如果不可能以用户可以成为发送者或接收者的方式对它们进行分组，那么也可以：

{_id: 2, sender:1, receiver:2, txt: “how are you 2”, date: “2018/04/02”}
{_id: 3, sender:2, receiver:1, txt: “hi 1”, date: “2018/04/03”}
{_id: 5, sender:1, receiver:3, txt: “how are you 3”, date: “2018/04/02”}

你们怎么看？可以用mongo / mongoose做到这一点吗？我当时在查看聚合和分组，但是对它们却不太熟悉。

Answer 1

您可以执行以下操作：

boolean x = true;