c# - 使用node对mongodb进行多次查询

Question

我似乎找不到我的问题的正确答案，我已经浏览了 stackoverflow。考虑一个 C# 示例代码，我试图将其移植到 Nodejs + mongobd。

var messageList = new List<MessageHelper>();

foreach( MessageActivation messageToAsk in messagesToAsk )
{
    var message     = from m in dbContext.Messages where m.Id == messageToAsk.MessageId select m;
    var jokeMessage = from m in dbContext.Messages where m.Id == messageToAsk.JokeMessageId select m;
    var user        = from u in dbContext.Users where u.Id == messageToAsk.SourceUserId select u;

    var messageHelper = new MessageHelper();
        messageHelper.AskingUserId = user.ToList()[0].Id;
        messageHelper.Message = message.ToList()[0];

    messageList.Add( messageHelper );
}

return messageList;

以 Node 的方式做到这一点的最佳方法是什么？只是一个侧 Node ，dbContext 将查询数据库。重点是收集所有需要的信息，将其打包然后发送。

谢谢

编辑:

这是我尝试过的

// collect needed info to make next queries in db
var messageIdList = new Array();
var jokeMessageIdList = new Array();
var sourceUserIdList  = new Array();
for( var i=0; i < messagesToAsk.length; i++ )
{
    messageIdList.push( messagesToAsk[i].MessageId      );
    jokeMessageIdList.push( jokeMessageId[i].JokeMessageId  );
    sourceUserIdList .push( jokeMessageId[i].SourceUserId   );
}

// make requests to have all the data in place
var messages = App.DataModels.Messages.find( {} );
    messages.where( 'MessageId' ).in( messageIdList );
    messages.exec( function ( err, foundMessages ) 
    {
        var jokeMessages = App.DataModels.Messages.find( {} );
            jokeMessages.where( 'JokeMessageId' ).in( jokeMessageIdList );
            jokeMessages.exec( function ( err, foundJokeMessages ) 
            {
                var users = App.DataModels.Messages.find( {} );
                    users.where( 'SourceUserId' ).in( sourceUserIdList );
                    users.exec( function ( err, foundUsers ) 
                    {
                        var messageList = new Array(); // new List<MessageHelper>();

                        for( var i=0; i < messagesToAsk.length; i++ )
                        {
                            var message     = null;
                            var jokeMessage = null;
                            var user        = null;

                            // get the data
                            for( var j = 0; j < messages.length; j++ )
                            {
                                if( messages[j].MessageId === messagesToAsk[i].MessageId )
                                {
                                    message = messages[j];
                                    break;
                                }
                            }
                            for( var k = 0; k < jokeMessages.length; k++ )
                            {
                                if( jokeMessages[k].JokeMessageId === messagesToAsk[k].JokeMessageId )
                                {
                                    jokeMessage = jokeMessage[k];
                                    break;
                                }
                            }
                            for( var l = 0; l < users.length; l++ )
                            {
                                if ( users[l].SourceUserId === messagesToAsk[l].SourceUserId )
                                {
                                    user = users[l];
                                    break;
                                }
                            }

                            var messageHelper = 
                            {
                                "AskingUserId"  : user.Id,
                                "AskingUserPic" : user.HelperPhoto,
                                "Message"       : message,
                                "JokeMessage"   : message.Type === "1" ? jokeMessage.Content
                            };

                            messageList.Add( messageHelper );
                        }

                        responseDelegate( response, messageList );
                    });
            });
    });

我把它贴在这里是为了其他正在轮类的人。这个想法似乎是，您需要先收集并过滤所有数据，然后再开始迭代和组装任何最终结果以发送。

Answer 1

在类比 NodeJS/Mongo 的 C#/Linq DB 查询的背景下回答我的问题。一种“思维”方式的映射。这主要是为了理解这种转变。正确的实现是使用像 Promise/Deffered 这样的模式(库)，如 @WiredPrairie 对问题的评论中所建议的。

// collect needed info to make next queries in db
var messageIdList = new Array();
var jokeMessageIdList = new Array();
var sourceUserIdList  = new Array();
for( var i=0; i < messagesToAsk.length; i++ )
{
    messageIdList.push( messagesToAsk[i].MessageId      );
    jokeMessageIdList.push( jokeMessageId[i].JokeMessageId  );
    sourceUserIdList .push( jokeMessageId[i].SourceUserId   );
}

// make requests to have all the data in place
var messages = App.DataModels.Messages.find( {} );
    messages.where( 'MessageId' ).in( messageIdList );
    messages.exec( function ( err, foundMessages ) 
    {
        var jokeMessages = App.DataModels.Messages.find( {} );
            jokeMessages.where( 'JokeMessageId' ).in( jokeMessageIdList );
            jokeMessages.exec( function ( err, foundJokeMessages ) 
            {
                var users = App.DataModels.Messages.find( {} );
                    users.where( 'SourceUserId' ).in( sourceUserIdList );
                    users.exec( function ( err, foundUsers ) 
                    {
                        var messageList = new Array(); // new List<MessageHelper>();

                        for( var i=0; i < messagesToAsk.length; i++ )
                        {
                            var message     = null;
                            var jokeMessage = null;
                            var user        = null;

                            // get the data
                            for( var j = 0; j < messages.length; j++ )
                            {
                                if( messages[j].MessageId === messagesToAsk[i].MessageId )
                                {
                                    message = messages[j];
                                    break;
                                }
                            }
                            for( var k = 0; k < jokeMessages.length; k++ )
                            {
                                if( jokeMessages[k].JokeMessageId === messagesToAsk[k].JokeMessageId )
                                {
                                    jokeMessage = jokeMessage[k];
                                    break;
                                }
                            }
                            for( var l = 0; l < users.length; l++ )
                            {
                                if ( users[l].SourceUserId === messagesToAsk[l].SourceUserId )
                                {
                                    user = users[l];
                                    break;
                                }
                            }

                            var messageHelper = 
                            {
                                "AskingUserId"  : user.Id,
                                "AskingUserPic" : user.HelperPhoto,
                                "Message"       : message,
                                "JokeMessage"   : message.Type === "1" ? jokeMessage.Content
                            };

                            messageList.Add( messageHelper );
                        }

                        responseDelegate( response, messageList );
                    });
            });
    });