我创建了一个（人员，学生，员工，教职员工）课程。人必须是学生和员工的子类。 员工有两个子类教职员工。除了我的驱动程序类 TestPerson程序 给出编译错误之外，我已经完成了所有编码工作，但它们正常工作
注意：一个测试程序，用于创建Person，Student，Employee，Faculty，Staff，并调用他们的toString方法。
驱动程序类 TestPerson.java 的错误如下： -
error: constructor Student in class Student cannot be applied to given types; error: no suitable constructor found for Employee(String,String,String,String) error: constructor Faculty in class Faculty cannot be applied to given types; error: no suitable constructor found for Staff(String,String,String,String)"
**我只是提供驱动程序类的代码。如果您需要我其他课程的其他编码，请在评论中说明，我会立即发布。
请参阅下面的编码： -
public class TestPerson {
public static void main(String[] args) {
Person person = new Person("John Doe", "123 Somewhere", "415-555-1212", "johndoe@somewhere.com");
Person student = new Student("Mary Jane", "555 School Street", "650-555-1212", "mj@abc.com", "junior");
Person employee = new Employee("Tom Jones", "777 B Street", "40-88-889-999", "tj@xyz.com");
Person faculty = new Faculty("Jill Johnson", "999 Park Ave", "92-52-22-3-333", "jj@abcxyz.com");
Person staff = new Staff("Jack Box", "21 Jump Street", "707-21-2112", "jib@jack.com");
System.out.println(person.toString() + "\n");
System.out.println(student.toString() + "\n");
System.out.println(employee.toString() + "\n");
System.out.println(faculty.toString() + "\n");
System.out.println(staff.toString() + "\n");
}
}
//人员类
public class Person {
private String name;
private String address;
private String phone_number;
private String email_address;
public Person() {
}
public Person(String newName, String newAddress, String newPhone_number, String newEmail){
name = newName;
address = newAddress;
phone_number = newPhone_number;
email_address = newEmail;
}
public void setName(String newName){
name = newName;
}
public String getName(){
return name;
}
public void setAddress(String newAddress){
address = newAddress;
}
public String getAddress(){
return address;
}
public void setPhone(String newPhone_number){
phone_number = newPhone_number;
}
public String getPhone(){
return phone_number;
}
public void setEmail(String newEmail){
email_address = newEmail;
}
public String getEmail(){
return email_address;
}
public String toString(){
return "Name :"+getName();
}
}
//学生班
public class Student extends Person {
public final String class_status;
public Student(String name, String address, int phone, String email, String classStatus) {
super(name, address, phone, email);
class_status = classStatus;
}
public String toString(){
return "Student Status: " + super.getName();
}
}
//员工类
import java.util.Date;
public class Employee extends Person{
private String office;
private double salary;
private Date hire;
public Employee() {
}
public Employee(String name, String address, int phone, String email){
super(name, address, phone, email);
}
public Employee(String office, double salary, Date hire){
this.office = office;
this.salary = salary;
this.hire = hire;
}
public void setOffice(String office){
this.office = office;
}
public String getOffice(){
return this.office;
}
public void setSalary(double salary){
this.salary = salary;
}
public double getSalary(){
return this.salary;
}
public void setHire(Date hire){
this.hire = hire;
}
public Date getHire(){
return this.hire;
}
public String toString(){
return "Office " + super.getName();
}
}
//教师班
public class Faculty extends Employee {
private String officeHours;
private int rank;
public Faculty(String name, String address, int phone, String email) {
super(name, address, phone, email);
}
public String toString(){
return "Office " + super.getOffice();
}
}
//职员班
public class Staff extends Employee {
private String title;
public Staff(String name, String address, int phone, String email) {
super(name, address, phone, email);
}
public Staff(String title){
this.title = title;
}
public void setTitle(String title){
this.title = title;
}
public String getTitle(){
return this.title;
}
public String toString(){
return "Title :" + super.getName();
}
}
答案 0 :(得分：4)
您获得这些错误的原因是构造函数不存在。
错误：构造函数类Student中的Student不能应用于给定 类型;错误：找不到合适的构造函数 雇员（字符串，字符串，字符串，字符串）
这意味着在你拥有这个代码之前你不会得到编译代码：
Student(String name, String addr, String phone, String email) {
....
}
假设您已在构造函数中设置了属性，toString将如下所示：
public String toString() {
return this.name + "\n" + this.addr + "\n" + this.phone + "\n" + this.email;
}
<强>更新
你的问题是学生只有这个构造函数：
public Student(String name, String address, int phone, String email, String classStatus)
学生需要一个构造函数，它只需要四个字符串作为参数。或者，您可以使所有内容都采用您指定的五个参数。
答案 1 :(得分：2)
它可能与问题本身无关，但我认为设计可以这样改进：
然后在Person和所有角色实现中定义toString
通过这种方式，您将能够独立于角色扩展或修改人员，从而使设计更加灵活
答案 2 :(得分：2)
Person的构造函数需要String作为第三个参数，但是您尝试将int phone传递给子类中的超级构造函数。这不会起作用，因为它的类型错误。
顺便说一下：你应该总是用字符串表示电话号码，而不是用整数表示。
答案 3 :(得分：1)
首先在所有类中将电话号码数据类型设置为整数..
主要功能将是：
public class TestPerson {
public static void main(String[] args) {
Person person = new Person("John Doe", "123 Somewhere", "415-555-1212",
"johndoe@somewhere.com");
Person student = new Student("Mary Jane", "555 School Street",
650-555-1212, "mj@abc.com", "junior");
Person employee = new Employee("Tom Jones", "777 B Street",
40-88-889-999, "tj@xyz.com");
Person faculty = new Faculty("Jill Johnson", "999 Park Ave",
92-52-22-3-333, "jj@abcxyz.com");
Person staff = new Staff("Jack Box", "21 Jump Street", 707-21-2112,
"jib@jack.com");
System.out.println(person.toString() + "\n");
System.out.println(student.toString() + "\n");
System.out.println(employee.toString() + "\n");
System.out.println(faculty.toString() + "\n");
System.out.println(staff.toString() + "\n");
}
}