比较列表中的整数

Question

嗨，我需要打印我的最高工资的人的方法。我有女巫雇员和女巫学生名单。当学生的平均成绩在4.5分以上时，他可以得到500英镑的薪水。

private void persons() {
    persons = new ArrayList<>();
    persons.add(new Employee("Pawel", "Kowalski", 2500));
    persons.add(new Employee("Kuba", "Nowak",  5500));
    persons.add(new Employee("Jacek", "Dudus",  2650));
    persons.add(new Employee("Jan", "Kot", 8000));
    persons.add(new Student("Ola", "Slowik", 4.42));
    persons.add(new Student("Kamil", "Slimak" 4.79));
    persons.add(new Student("Emil", "Skalski", 3.75));
    persons.add(new Student("Ula", "Mnich", 4.85));

我有这样方法：

private Person  bestPayed(List<Person> persons){
    persons.sort(Comparator.comparingInt(Person::getButHereICantFindGetSalary));

return
}

这是Person类：

public class Person {
private String name;
private String surname;
private int age;

public Person(String name, String surname, int age) {
    this.name = name;
    this.surname = surname;
    this.age = age;
}

public String getName() {
    return name;
}

public String getSurname() {
    return surname;
}

public int getAge() {
    return age;
}

这是学生课堂：

public class Student extends Person {
private String fieldOfStudy;
private double averageGrade;


public Student(String name, String surname, int age, String fieldOfStudy, double averageGrade) {
    super(name, surname, age);
    this.fieldOfStudy = fieldOfStudy;
    this.averageGrade = averageGrade;
}

public String getFieldOfStudy() {
    return fieldOfStudy;
}

public double getAverageGrade() {
    return averageGrade;
}

我需要的方法为最好的支付的人和方法总和所有人的收入。

Answer 1

“GetBuThereIcantFindGetSalary”

您找不到getSalary()，因为您还没有创建。

您需要向person类添加一个新方法getSalary():

public abstract class Person {
…
public abstract int getSalary();
…
}

然后在学生和雇主中实施：

public class Student extends Person {
…
public int getSalary() {
    return /*Calculate salary here*/;
}
…
}

public class Employer extends Person {
…
public int getSalary() {
    return salary;
}
…
}