如何更好地優化這些數據庫查詢/功能流？

Question

我已將手從游戲開發轉向編寫支持后端。 到目前為止，一切似乎都有效，但是當我開始編寫朋友系統時，我遇到了這個功能流程，這對我來說似乎很臟。 而且我確定我現在只是在一起攻擊它。 任何node.js向導都要告訴我如何改進這個？

相當肯定我應該在Redis中緩存玩家查找。

acceptfriend: function(req, res){
    //Find our user
    User.findById( req.decoded._id, function(err, user){
        //error occured
        if(err){
            return res.status(401).send(err);
        }
        //no user found
        if(!user){
            return res.status(401).json({
                succes: false,
                message: 'no user found with that id'
            } );
        }
        //Does the request exist?
        if( !_.any( user.social.friendRequests, {id: req.params.id} ) ){
            return res.status(401).json( {
                succes: false,
                message: 'friend request not found'
            } );
        }
        //find the user that belongs to the request
        User.findById( req.params.id, function(err, friend){
            //error occured
            if(err){
                return res.send(err);
            }
            //user doesnt exist
            if(!friend){
                return res.status(401).json({
                    succes: false,
                    message: 'no user found with that id'
                } );
            }
            //Pull the request from the friendRequests array
            user.social.friendRequests.pull( req.params.id );
            //Add the friend
            user.social.friends.addToSet( {
                user_id: friend._id,
                name: friend.username,
                corp: 'n/a'
            } );
            //Add the user to the friends list as well
            friend.social.friends.addToSet({
                user_id: user._id,
                name: user.username,
                corp: 'n/a'
            });
            //save the docs
            user.save();
            friend.save();
        } );
        //return success
        return res.status(200).json({
            success: true,
            message: 'friend succesfully added'
        });
    } );
}

Answer 1

1-首先，你有一個很大的功能。 你必須將它拆分成一些功能。 這樣做可以使用任何測試框架測試它們。

2-將錯誤響應句柄委派給控制器。

from -> return res.status(401).send(err);
to (with Promises)-> deferred.reject(err); 
to (normal way)   -> throw new Error(err); 

3-您可以使用Promises來管理節點的異步行為以清除代碼。 我創建了一個例子，也許是第一次沒有工作，隨時修復不正確的引用。 用戶參考，'接受朋友'方法......

要點： https ： //gist.github.com/aitoraznar/b7099ad88ead0cdab256

 var Promise = require('bluebird'); var _ = require('lodash'); //var User = app.models.User; var ERRORS = { userNotFoundError: { code: 401, success: false, message: 'no user found with that id' }, friendRequestNotFoundError: { code: 401, success: false, message: 'friend request not found' }, friendNotFoundError: { code: 401, success: false, message: 'no friend found with that id' } } var SUCCESS_MESSAGES= { friendAddedSuccessfully: { success: true, message: 'friend succesfully added' } }; var userDAO = { /* * */ getUserById: function(id) { var deferred = Promise.pending(); User.findById(id, function(err, user) { //error occured if (err) { err.code = 401; return deferred.reject(err); } //no user found if (!user) { return deferred.reject(ERRORS.userNotFoundError); } deferred.resolve(user); }); return deferred.promise; }, /* * Does the request exist? */ checkFriendRequest: function(user, friendId) { var deferred = Promise.pending(); if (userDAO.haveFriendRequestFrom(user, friendId)) { deferred.resolve(user, friendId); } else { return deferred.reject(ERRORS.friendRequestNotFoundError); } return deferred.promise; }, /* * */ haveFriendRequestFrom: function(user, friendId) { return _.any(user.social.friendRequests, {id: friendId }); }, /* * */ getFriend: function(user, friendId) { var deferred = Promise.pending(); userDAO.getUserById(friendId) .then(function(friend) { deferred.resolve(user, friend); }, function(error) { if (error === ERRORS.userNotFoundError) { // Then the error is friend not found // Override the error error = ERRORS.friendNotFoundError; } return deferred.reject(error); }); return deferred.promise; }, /* * */ makeFriendship: function(user, friend) { var deferred = Promise.pending(); //Pull the request from the friendRequests array user.social.friendRequests.pull(friend._id); //Add the friend user.social.friends.addToSet( { user_id: friend._id, name: friend.username, corp: 'n/a' } ); //Add the user to the friends list as well friend.social.friends.addToSet({ user_id: user._id, name: user.username, corp: 'n/a' }); //save the docs user.save(); friend.save(); // Return the new friendship var friendship = { user: user, friend:friend }; deferred.resolve(friendship); return deferred.promise; }, /* * */ friendRequestError: function(err) { var deferred = Promise.pending(); // Propagate de error deferred.reject(err); return deferred.promise; }, /* * */ friendRequest: function(userId, friendId) { var deferred = Promise.pending(); // Get user by ID userDAO.getUserById(userId) // Check if the user is able to add the friend .then(userDAO.checkFriendRequest, userDAO.friendRequestError) // Get the friend to add .then(userDAO.getFriend, userDAO.friendRequestError) // Make the friendship .then(userDAO.makeFriendship, userDAO.friendRequestError) // Response to the controller .then( function(friendship) { // Resolve with new friendship // This goes to 'success' function in controller deferred.resolve(friendship); }, function(error) { // This goes to 'error' function in controller deferred.reject(error); }) return deferred.promise; } }; // Controller var acceptfriend = function(req, res, next) { var userId = req.decoded._id; var friendId = req.params.id; userDAO.friendRequest(userId, friendId) .then(function(friendRequest) { console.log('---> SUCCESS'); //return success return res.status(200) .json(SUCCESS_MESSAGES.friendAddedSuccessfully); }, function(error) { console.error('---> ERROR', error); return res.status(error.code).json(error); }); } 

4-在集合/表中創建數據庫索引

此致，Aitor