使用不同参数创建类似子类实例的方法

时间:2018-05-30 11:51:15

标签: java methods refactoring subclass

我目前正在学习Java。我有一个带超类(IndexCard)的项目,有两个子类(EmployeeIndexCard和CustomerIndexCard)。两个子类非常相似,但它们的实例变量彼此不同,因此它们的构造函数也不同。

他们是:

class EmployeeIndexCard extends IndexCard {
    public WorkArea workArea ;
    protected String password;

    public employeeIndexCard(String name, String password, String adress, String phone, String email, String workArea) {
        super(name, adress, phone, email);
        this.password = password;
        this.workArea = WorkArea.valueOf(workArea);
    }
}


class CustomerIndexCard extends IndexCard {
    public customerIndexCard(String name, String adress, String phone, String email) {
        super(name, adress, phone, email);
    }
}

我想知道我做错了什么,因为要创建这些类的实例,我创建了两个非常相似的方法:

/**
 * Create an instance of EmployeeIndexCard.
 */
public static void employeeIndexCard(String name, String dni, String password, String adress, String phone, String email, String workArea) {
    if (Utils.validateDni(dni) && !IndexCard.list.containsKey(dni)) {
        IndexCard.list.put(dni, new EmployeeIndexCard(name, password, adress, phone, email, workArea));
    } else {
        throw new InvalidParameterException();
    }
}

/**
 * Create an instance of CustomerIndexCard.
 */
public static void customerIndexCard(String name, String dni, String adress, String phone, String email) {
    if (Utils.validateDni(dni) && !IndexCard.list.containsKey(dni)) {
        IndexCard.list.put(dni, new FichaCliente(name, adress, phone, email));
    } else {
        throw new InvalidParameterException();
    }
}

有没有办法重构代码以合并这两个几乎完全相同的方法?

2 个答案:

答案 0 :(得分:2)

由于您的两个类共享一个父类,因此重构代码的最自然方式是负责为调用者创建实例,并接受IndexCard类型的任何实例:

public static void addIndexCard(String dni, IndexCard indexCard) {
    if (Utils.validateDni(dni) && !IndexCard.list.containsKey(dni)) {
        IndexCard.list.put(dni, indexCard);
    } else {
        throw new InvalidParameterException();
    }
}

这样,您可以简单地调用它:

//add customer index card:
addIndexCard("dni", new FichaCliente(name, adress, phone, email));

//add employee index card:
addIndexCard("dni2", new EmployeeIndexCard(name, password, adress, 
             phone, email, workArea));

答案 1 :(得分:1)

从我的角度来看,问题是你在这里反对面向对象的设计。您没有提供对IndexCard类的访问权限,但它看起来应该是这样的:

public class IndexCard {

    public static Map<String, IndexCard> map = new HashMap<>();

    private String name;
    private String address;
    private String phone;
    private String email;

    // constructor and accessors ommitted
}

首先,请不要使用public static字段,并在需要时提供访问者。这可以防止其他类直接更改其状态。您还可以将验证逻辑放在那里:

public class IndexCard {

    private static Map<String, IndexCard> map = new HashMap<>();

    public static void addIndexCard(String dni, IndexCard card) {
        if (Utils.validateDni(dni) && !map.containsKey(dni)) {
            map.put(dni, card);
        } else {
            throw new InvalidParameterException();
        }
    }

    private String name;
    private String address;
    private String phone;
    private String email;

    // constructor and accessors ommitted
}

您可以使用此类

IndexCard c1 = new EmployeeIndexCard(name, password, adress, phone, email, workArea);
IndexCard.addIndexCard("c1", c1);
IndexCard c2 = new FichaCliente(name, adress, phone, email);
IndexCard.addIndexCard("c2", c2);