Question

我已经通过MongoDB进行了尝试，但是我无法在mongoDB中使用JOIN Query，并且我的项目足够广泛。因此，要在MySQL的节点js中创建Restful API。谁能建议解决方案

Answer 1

对于带有NodeJS的MySQL，您可以使用Sequelize，这有点像symfony中的ORM学说 http://docs.sequelizejs.com/

Answer 2

要创建REST API，您可以使用Express JS

var express = require('express');
var app = express();    

app.get('/', function (req, res) {
   //BELOW-CODE
});

您可以通过以下代码连接Mysql：

var mysql = require('mysql');

var con = mysql.createConnection({
 host: "localhost",
 user: "yourusername",
 password: "yourpassword",
 database: "mydb"
});

con.connect(function(err) {
 if (err) throw err;
 con.query("SELECT * FROM customers", function (err, result, fields) {
 if (err) throw err;
 console.log(result);
    );
  });

注意：安装expressJS框架以开始快乐的编码：-）

Answer 3

 `'user strict';
var sql = require('./db.js');

//Task object constructor
var Task = function(task){
    this.task = task.task;
    this.status = task.status;
    this.created_at = new Date();
};
Task.createTask = function createUser(newTask, result) {    
        sql.query("INSERT INTO tasks set ?", newTask, function (err, res) {

                if(err) {
                    console.log("error: ", err);
                    result(err, null);
                }
                else{
                    console.log(res.insertId);
                    result(null, res.insertId);
                }
            });           
};
Task.getTaskById = function createUser(taskId, result) {
        sql.query("Select task from tasks where id = ? ", taskId, function (err, res) {             
                if(err) {
                    console.log("error: ", err);
                    result(err, null);
                }
                else{
                    result(null, res);

                }
            });   
};
Task.getAllTask = function getAllTask(result) {
        sql.query("Select * from tasks", function (err, res) {

                if(err) {
                    console.log("error: ", err);
                    result(null, err);
                }
                else{
                  console.log('tasks : ', res);  

                 result(null, res);
                }
            });   
};
Task.updateById = function(id, task, result){
  sql.query("UPDATE tasks SET task = ? WHERE id = ?", [task.task, id], function (err, res) {
          if(err) {
              console.log("error: ", err);
                result(null, err);
             }
           else{   
             result(null, res);
                }
            }); 
};
Task.remove = function(id, result){
     sql.query("DELETE FROM tasks WHERE id = ?", [id], function (err, res) {

                if(err) {
                    console.log("error: ", err);
                    result(null, err);
                }
                else{

                 result(null, res);
                }
            }); 
};

module.exports= Task;

使用MySQl的NodeJs Restful Api

3 个答案: