如何对扩展类的 ArrayList 进行排序？

Question

这是一个主类Employee.java 和另外两个扩展主类的类。 ArrayList<Employee> 包含来自 createFixedEmployee.java 和 createPerHourEmployee 的对象。如何将ArrayList中的对象按薪水排序，如果某些对象的薪水相同，则按字母顺序对其名称进行排序？

我尝试使用Comparator.comparing(); 但不起作用，我收到错误Cannot resolve method getMonthSalary(); 这是代码：

createPerHourEmployee.java

public class createPerHourEmployee extends Employee {
    private double salary;

    public createPerHourEmployee() {}

    public double getSalaryPerHour() { return this.salary; }

    public double setSalaryPerHour(double value) {
        return this.salary = value;
    }


    public createPerHourEmployee (int _id, String _name, double _salary) {
        setEmployeeID(_id);
        setEmployeeName(_name);
        this.salary = _salary;
    }

    public double getMonthSalary() {
        return salary * (20 * 0.8);
    }

    public String toString() {
        return getEmployeeID() + ", " + getEmployeeName() + ", " + getSalaryPerHour();
    }
}

createFixedEmployee.java

public class createFixedEmployee extends Employee {
    private double salary;

    public createFixedEmployee() {}

    public double getSalaryFixed() {
        return this.salary;
    }

    public double setSalaryFixed(double value) {
        return this.salary = value;
    }

    public createFixedEmployee(int _id, String _name, double _salary) {
        setEmployeeID(_id);
        setEmployeeName(_name);
        this.salary = _salary;
    }

    public double getMonthSalary() {
        return this.salary;
    }

    public String toString() {
        return getEmployeeID() + ", " + getEmployeeName() + ", " + getSalaryFixed();
    }

}

Employee.java

import java.util.ArrayList;
import java.util.Comparator;

public abstract class Employee {
    private int base_id;
    private String base_name;

    public int getEmployeeID () {
        return this.base_id;
    }

    public int setEmployeeID (int value) {
        return this.base_id = value;
    }

    public String getEmployeeName () {
        return this.base_name;
    }

    public String setEmployeeName(String value) {
        return this.base_name = value;
    }

    public abstract double getMonthSalary();

    public static void main(String[] argc) {
        ArrayList<Employee> Employee = new ArrayList<Employee>();
        Employee.add(new createPerHourEmployee(1, "asd", 1300));
        Employee.add(new createFixedEmployee(7, "asds", 14025));
        Employee.add(new createPerHourEmployee(2, "nikan", 1230));
        Employee.add(new createPerHourEmployee(3, "nikalo", 12330));
        Employee.add(new createFixedEmployee(6, "aaaa", 14025));
        Employee.add(new createFixedEmployee(4, "nikaq", 140210));
        Employee.add(new createFixedEmployee(5, "nikas", 124000));
        Employee.add(new createFixedEmployee(6, "nikab", 14025));


        Employee.sort(Comparator.comparing(Employee::getMonthSalary)
              .thenComparing(Employee::getEmployeeName)); // here is an error

    }
}

所以我尝试过使用

Employee.sort(Comparator.comparing(createFixedEmployee::getMonthSalary).
                thenComparing(createFixedEmployee::getEmployeeName));

效果很好，但我需要对所有类进行排序，而不仅仅是一个类。

Answer 1

在main 中，您为局部变量指定了名称Employee。这会影响您的 Employee 类的标识符。变量名无论如何都不应该以大写字母开头，应该在引用数组或列表时使用复数形式（除非它们以“list”或类似名称结尾），理想情况下不应该是与您的任何班级名称相同。

因此将变量更改为employees。修复这个问题，使用Employee::getMonthSalary 的版本和这样的作品：

employees.sort(Comparator.comparing(Employee::getMonthSalary)
      .thenComparing(Employee::getEmployeeName));

---

我还将createPerHourEmployee 更改为PerHourEmployee，因为Java 中压倒性的约定是类名以大写字母开头并且是名词，而不是动词。同样，createFixedEmployee 将是（比如说）SalariedEmployee。

Answer 2

当对List<T> 进行排序时，您绝对可以使用对T 的任何祖先方法的函数引用。这里的问题在于您为列表提供的名称 - 您已将局部变量命名为 Employee，当您尝试在 Comparator 中创建函数引用时，它隐藏了类型 Employee 的名称。只要给它一个更好的名字，你应该没问题。例如：

ArrayList<Employee> employeeList = new ArrayList<Employee>();
// Or better yet: List<Employee> employeeList = new ArrayList<>();

employeeList.add(new createPerHourEmployee(1, "asd", 1300));
employeeList.add(new createFixedEmployee(7, "asds", 14025));
employeeList.add(new createPerHourEmployee(2, "nikan", 1230));
employeeList.add(new createPerHourEmployee(3, "nikalo", 12330));
employeeList.add(new createFixedEmployee(6, "aaaa", 14025));
employeeList.add(new createFixedEmployee(4, "nikaq", 140210));
employeeList.add(new createFixedEmployee(5, "nikas", 124000));
Employee.add(new createFixedEmployee(6, "nikab", 14025));


employeeList.sort(Comparator.comparing(Employee::getMonthSalary)
      .thenComparing(Employee::getEmployeeName));