Mongoose 如何获取对话列表，特定用户与之聊天？

Question

好的，我的数据模型基本上是这样的：

var messageSchema = new Schema({
    to: { type: String, required: true},
    from: { type: String, required: true},
    message: { type: String, required: true}
});

我想做的是有一个功能，我可以传入当前登录用户的用户名：例如 billy。

数据库可能包含 100 条这样的消息：

[{"_id":"2394290384","from":"billy","to":"dan","message":"some message"},
 {"_id":"2394290384","from":"dan","to":"billy","message":"some message"},
 {"_id":"2394290389","from":"john","to":"billy","message":"some message"},
 {"_id":"2394290389","from":"billy","to":"john","message":"some message"}] 

它应该只给我这样的输出：

 [{"_id":"2394290384","from":"billy","to":"dan","message":"some message"},   
  {"_id":"2394290389","from":"john","to":"billy","message":"some message"}] 

我怎样才能为每条消息提取最新的 1。

因此，无论是我向他们发送消息还是他们向我发送消息，它基本上都会显示我与之聊天的每个对话/每个人的 1 个结果。

每个对话应该只提取 1 个结果，其中包含与其最新消息相关的所有数据。

所以最新的消息，该消息的 id，它的来往用户。

这样我就可以在一侧创建一个对话列表，在用户名下方显示用户及其消息。用户点击它，它会显示聊天。

我已经问过这个问题的多种变体，试图弄清楚，但到目前为止我还没有成功地找到解决方案。

我来自 php mysql 背景，所以我是 node 和 mongoose 的新手。

提前感谢您的所有帮助，我真的很感激。

我已经提到了为什么示例中的 2 个项目是不同的。根据当前登录的用户名，每次对话都会提取 1 个结果。

所以需要有这样的函数：

getconversations(username){
Message.find(Where messages are to this username or from this from username).exec(function(error,results){
this would probably output all messages like so:
[{"_id":"2394290384","from":"billy","to":"dan","message":"some message"},
 {"_id":"2394290384","from":"dan","to":"billy","message":"some message"},
 {"_id":"2394290389","from":"john","to":"billy","message":"some message"},
 {"_id":"2394290389","from":"billy","to":"john","message":"some message"}] 


})
}

if the current username was billy.

    But I only want 1 of the latest.

    So i can have a list on the left:
    messages:
    username:
    lastmessage
    username:
    last message

    a user clicks on it and it opens up the full chat. For example like you see on skype on ios or android: it shows you a list of users and their message below. you tap on it and it shows you the chat. it is basically pulling that list based on people you have had a conversation with.

我最终是如何让它工作的：

var user="billy";
Message.aggregate(
    {
           $match: {
              $or: [{
                to: user
              }, 
              {
                from: user
              }]
            }
        },
    { $project: { "from" : {
      $cond: {'if': {$eq: ['$to',user]},'then': '$from', 'else': '$to'}
                            },
                            "to":{
      $cond: {'if': {$eq: ['$to',user]},'then': '$to', 'else': '$from'}
                            },
                            "message":"$message"



                } },
    { $sort: { _id: -1 } },
    { $group: { "id": { "$first" : "$_id" },"_id" : { "from" : "$from" }, "from": { "$first" : "$from" },"to": { "$first" : "$to" }, "message": { "$first" : "$message" }  } }
, function(err, docs) {
  res.send(docs);

});

它按特定的用户名搜索并删除所有重复项，只为每个用户输出 1 条记录，并输出他们的消息、ID、来往和来自

Answer 1

我假设问题是您如何在 Mongoose 的 Mongodb 中执行 or 查询？

var Message = mongoose.model('message');
Message.find({ $or: [ { from: 'billy' }, { to: 'billy' } ] }, function(err, docs) {
    // docs will now be an array of Message models
});

更新： 您可以根据插入时间按_id 排序，然后将结果限制为一条记录：

var Message = mongoose.model('message');
Message.find({ $or: [ { from: 'billy' }, { to: 'billy' } ] })
    .sort({ _id : -1 })
    .limit(1)
    .exec(function(err, docs) {
    // docs will now be an array containing 1 Message model
});

更新 2：
尝试使用聚合框架：

var Message = mongoose.model('message');
Message.aggregate(
    { $project: { "from" : "$from", "message" : "$message" } },
    { $sort: { _id: -1 } },
    { $group: { "_id" : { "from" : "$from" }, "from": { "$first" : "$from" }, "message": { "$first" : "$message" } } }
, function(err, docs) {
    /* i.e. docs = array
    [ 
        { "_id" : { "from" : "alex" }, "from" : "alex", "message" : "Some message" },
        { "_id" : { "from" : "billy" }, "from" : "billy", "message" : "Some message" },
    ...

    */
});