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Design Pattern Tutorials

Factory Design Pattern Introduction

What is Factory Design Pattern Gang of Four Definition : “Define an interface for creating an object, but let sub-classes decide which class to instantiate. The Factory method lets a class defer instantiation it uses to sub-classes”

Factory pattern is one of the most used design patterns in real world applications

Factory pattern creates object without exposing the creation logic to the client and refer to newly created object using a common interface


From the above diagram, client uses factory and creates the product.

Implementation Guidelines : We need to choose Factory Pattern when The Object needs to be extended to subclasses The Classes doesn’t know what exact sub-classes it has to create The Product implementation tend to change over time and the Client remains unchanged Simple Factory Example : Business Requirement

Differentiate employees as permanent and contract and segregate their pay scales as well as bonus based on their employee types
We can address the above requirement with the below implementations

  • Implement without Factory Pattern
  • Use a Simple Factory
  • Enhance Simple factory to Factory Method Pattern
We will be working on the Employee Portal that we used in the Singleton tutorials. Please refer to them before proceeding.

Prerequisite steps
Step 1 : Enhance the DB model to add Employee_Type Table 
CREATE TABLE [dbo].[Employee_Type]
(
    [Id]           INT           IDENTITY (1, 1) NOT NULL,
    [EmployeeType] VARCHAR (150) NOT NULL,
    PRIMARY KEY CLUSTERED ([Id] ASC)

)

Step 2 : Add Permanent and Contract Employees as Master Data
Step 3 : Add new columns EmployeeTypeID, Bonus, HourlyPay to Emplyee Table and add Foreign key constraint to the Emp 
CREATE TABLE [dbo].[Employee]
(
    [Id]             INT          IDENTITY (1, 1) NOT NULL,
    [Name]           VARCHAR (50) NOT NULL,
    [JobDescription] VARCHAR (50) NOT NULL,
    [Number]         VARCHAR (50) NOT NULL,
    [Department]     VARCHAR (50) NOT NULL,
    [HourlyPay]      DECIMAL (18) NOT NULL,
    [Bonus]          DECIMAL (18) NOT NULL,
    [EmployeeTypeID] INT          NOT NULL,
    PRIMARY KEY CLUSTERED ([Id] ASC),
    CONSTRAINT [FK_Employee_EmployeeType] FOREIGN KEY ([EmployeeTypeID]) REFERENCES [dbo].[Employee_Type] ([Id])

Step 4 : Update the Emloyee Model edmx file with the latest changes
Step 5 : Create new BaseController and Move the Singleton Exception logic to the base controller 
public class BaseController : Controller
    {
        private ILog _ILog;
        public BaseController()
        {
            _ILog = Log.GetInstance;
        }
        protected override void OnException(ExceptionContext filterContext)
        {
            _ILog.LogException(filterContext.Exception.ToString());
            filterContext.ExceptionHandled = true;
            this.View("Error").ExecuteResult(this.ControllerContext);
        }
    }

Step 6 : Regenerate the EmployeesController and its corresponding views
Step 7 : Comment the code in Create and Update views which accepts inputs for Bonus and HourlyPay
Solution 1: Implement without Factory Pattern
EmployeeController.cs 
  [HttpPost]
        [ValidateAntiForgeryToken]
        public ActionResult Create([Bind(Include = "Id,Name,JobDescription,Number,Department,HourlyPay,Bonus,EmployeeTypeID")] Employee employee)
        {
            if (ModelState.IsValid)
            {
                if (employee.EmployeeTypeID == 1)
                {
                    employee.HourlyPay = 8;
                    employee.Bonus = 10;
                }
                else if (employee.EmployeeTypeID == 2)
                {
                    employee.HourlyPay = 12;
                    employee.Bonus = 5;
                }
                db.Employees.Add(employee);
                db.SaveChanges();
                return RedirectToAction("Index");
            }

            ViewBag.EmployeeTypeID = new SelectList(db.Employee_Type, "Id", "EmployeeType", employee.EmployeeTypeID);
            return View(employee);
        }

The above code introduces a) Tight coupling between Controller class and Business logic
b) For any new employee type addition, we end up modifying the controller code adding extra over heads in the development and testing process

Using a simple factory eliminates the above drawbacks.
Solution 2: Implement with Simple Factory
Step 1 : Add new Manager folder and add the below interface and classes

IEmployeeManager.cs 
 public interface IEmployeeManager
    {
        decimal GetBonus();
        decimal GetPay();
    }


ContractEmployeeManager.cs 
 public class ContractEmployeeManager : IEmployeeManager
    {
        public decimal GetBonus()
        {
            return 5;
        }

        public decimal GetPay()
        {
            return 12;
        }
    }


PermanentEmployeeManager.cs 
public class PermanentEmployeeManager : IEmployeeManager
    {
        public decimal GetBonus()
        {
            return 10;
        }

        public decimal GetPay()
        {
            return 8;
        }
    }

Step 2 : Create Factory folder and add the below Manager class
EmployeeManagerFactory.cs 
 public class EmployeeManagerFactory
    {
        public IEmployeeManager GetEmployeeManager(int employeeTypeID)
        {
            IEmployeeManager returnValue = null;
            if (employeeTypeID == 1)
            {
                returnValue = new PermanentEmployeeManager();
            }
            else if (employeeTypeID == 2)
            {
                returnValue = new ContractEmployeeManager();
            }
            return returnValue;
        }
    }

Step 3 : Update the employee’s controller to consume the factory. 

 [HttpPost]
        [ValidateAntiForgeryToken]
        public ActionResult Create([Bind(Include = "Id,Name,JobDescription,Number,Department,HourlyPay,Bonus,EmployeeTypeID")] Employee employee)
        {
            if (ModelState.IsValid)
            {
                EmployeeManagerFactory empFactory = new EmployeeManagerFactory();
                IEmployeeManager empManager = empFactory.GetEmployeeManager(employee.EmployeeTypeID);
                employee.Bonus = empManager.GetBonus();
                employee.HourlyPay = empManager.GetPay();
                db.Employees.Add(employee);
                db.SaveChanges();
                return RedirectToAction("Index");
            }

            ViewBag.EmployeeTypeID = new SelectList(db.Employee_Type, "Id", "EmployeeType", employee.EmployeeTypeID);
            return View(employee);
        }

;