Nodejs，返回條件不起作用

Question

我是Nodejs的新手。我正在嘗試返回數據。選擇查詢後。在這裏，我寫了兩個條件。首先是非常簡單的邏輯，它正在工作，但讓我知道爲什麼第二個條件不起作用。

第一個條件：

var arr = {email:"john@gmail.com", password:"}; 
return arr; 

databaseConnection.query("SELECT * FROM users where email = '"+email+"' and password = '"+password +"'", function (err, result) { 

}); 

第二個條件

databaseConnection.query("SELECT * FROM users where email = '"+email+"' and password = '"+password +"'", function (err, result) {  

var arr = {email:"john@gmail.com", password:"}; 
return arr; 

}); 

從passport.js

var LocalStrategy = require('passport-local').Strategy; 
var FacebookStrategy = require('passport-facebook').Strategy; 
var TwitterStrategy = require('passport-twitter').Strategy; 
var configAuth = require('./auth'); 


module.exports = function(passport,databaseConnection) { 

    var usermodule = require('../models/user')(databaseConnection); 

passport.serializeUser(function(user, done) { 
     done(null, user); 
    }); 

passport.deserializeUser(function(user, done) { 
     done(null, user); 
    }); 

passport.use('local-login', new LocalStrategy({ 

     usernameField : 'username', 
     passwordField : 'password', 
     passReqToCallback : true // allows us to pass back the entire request to the callback 

    }, 
    function(req, email, password, done) { // callback with email and password from our form 

      var user={}; 


       var result = usermodule.login(email,password); 

        console.log('usercraeted'); 
        user["email"]=email; 
        user["status"]=true; 
        user["name"]="John Snow"; 
        user["avatar"]=""; 
        user["loginStatus"]=true; 
        return done(null, user); 

       user["msg"]="invalide email"; 
       console.log("out"); 


      return done(null, false,user["msg"]); 
     }; 

    ); 

}; 

Answer 1

在你的第一個例子中，代碼被執行了這一點：

var arr = {email:"john@gmail.com", password:"}; 
return arr; 

// interpreter exits and `{email:"john@gmail.com", password:"}` is returned, 
// `databaseConnection.query("` is never executed 

在第二個例子中，代碼執行到了這一點：

databaseConnection.query("SELECT * FROM users where email = '"+email+"' and password = '"+password +"'", function (err, result) {  

var arr = {email:"john@gmail.com", password:"}; 
return arr; 

}); 
// interpreter exits, undefined is returned 

Answer 2

在S第二個條件不起作用，因爲node.js是異步的。而你的回報是需要一段時間才能執行的回調。

Answer 3

你必須使用回調。

function getUser() { 
    var email = "test@test.com"; 
    var password = "test"; 
    getDatas(email, password, function(err, result) { 
     // the result is good and not undefined if no errors :) 
    }); 
} 

function getDatas(email, password, callback) { 

    databaseConnection.query("SELECT * FROM users where email = '"+email+"' and password = '"+password +"'", function (err, result) {  
     callback(err, result); // callback is executed with the result you 
    }); 
} 

希望我幫你