如何仅从所需的类（流）获取数据？

Question

Employee class:

public abstract class Employee extends Person {
    private final Manager manager;
    private final BigDecimal salary;
    
    protected Employee(String firstName, String surname, LocalDate birth_date, Manager _manager, BigDecimal _salary) {
        super(firstName, surname, birth_date);
        manager = _manager;
        salary = _salary;
        if (manager != null) {
            manager.getSubordinates().add(this);
        }
    }

    ...
}

Worker class:

public class Worker extends Employee {
    private final LocalDate employment_date;
    private BigDecimal bonus;
    
    public Worker(String firstName, String surname, LocalDate birth_date, Manager manager, BigDecimal salary,
                  LocalDate _employment_date, BigDecimal _bonus) {
        super(firstName, surname, birth_date, manager, salary);
        employment_date = _employment_date;
        bonus = _bonus;
    }

    ...
}

Manager class:

public final class Manager extends Worker {
    List<Employee> subordinates = new ArrayList<Employee>();
    
    public Manager(String firstName, String surname, LocalDate birth_date, Manager manager, BigDecimal salary,
                   LocalDate employment_date, BigDecimal bonus) {
        super(firstName, surname, birth_date, manager, salary, employment_date, bonus);
    }

    ...
}

Trainee class:

public class Trainee extends Employee {

    private final LocalDate start_date;
    private final short apprenticeship_length;
    
    public Trainee(String firstName, String surname, LocalDate birth_date, Manager manager, BigDecimal salary,
                   LocalDate _start_date, short _apprenticeship_length) {
        super(firstName, surname, birth_date, manager, salary);
        manager.getSubordinates().add(this);
        start_date = _start_date;
        apprenticeship_length = _apprenticeship_length;
    }
}

payrol class:

public final class PayrollEntry {

    private final Employee _employee;
    private final BigDecimal _salaryPlusBonus;
    
    public PayrollEntry(Employee employee, BigDecimal salary, BigDecimal bonus) {
        _employee = employee;
        _salaryPlusBonus = salary.add(bonus);
    }
}

我必须编写函数 List<PayrollEntry> payroll(List<Employee> employees) {}。正如你在上面看到的，只有 Worker 和 Manager 可以有奖金，另一方面 Trainee 没有，但他们都来自 class Employee（顺便说一句，我无法更改 classes 层次结构中的任何内容，因为这是我的作业，层次结构是由老师编写的）。我应该使用函数式编程技术来编写函数，这是我的尝试：

public static List<PayrollEntry> payroll(List<Employee> employees) {
    return employees
            .stream()
            .map(employee -> new PayrollEntry(employee, employee.getSalary(), ((Worker) employee).getBonus()))
            .collect(Collectors.toList());
}

我明白为什么它会给我 ClassCastException，但我不知道使用 stream 的任何其他方法。我想我可以用 for-each 循环检查它是否是 Trainee 来完成它，但我想知道是否有使用 stream.

Answer 1

对 map() 的内容有什么限制吗？

否则你可以有类似的东西：

.map(employee -> {
    if (employee instanceof Worker) {
        return new PayrollEntry()....
    } else {
        return new PayrollEntry()....
    }
})

Answer 2

我不确定它是否符合你老师给你的任务定义，但你可以按如下方式扩展你的 map 流操作：

public static List<PayrollEntry> payroll(List<Employee> employees) {
        return employees
                .stream()
                .map(employee -> {
                    // if statements to check type of employee
                    // set some variables for the various fields
                    return new PayroleEntry(...);
                })
                .collect(Collectors.toList());
}

Answer 3

您可以在转换之前使用三元运算符并检查员工是否是 Worker class 的实例。如果是，则通过奖金 else BigDecimal.ZERO:

employees.stream()
         .map(employee -> new PayrollEntry(employee, employee.getSalary(),
               employee instanceof Worker ? ((Worker) employee).getBonus() : BigDecimal.ZERO))
         .collect(Collectors.toList());

Answer 4

我会为这些情况创建一个静态构造函数。（类似于 PayrollEntry.forEmployee(Employee)）因为你不应该改变原来的类你可以把这个方法放在其他地方。

private static PayrollEntry newPayrollEntry(Employee employee) {
    BigDecimal bonus = BigDecimal.ZERO;        
    if (employee instanceof Worker) {
        bonus = ((Worker) employee).getBonus();
    }
    return new PayrollEntry(employee, employee.getSalary(), bonus);
}

public static List<PayrollEntry> payroll(List<Employee> employees) {
    return employees
            .stream()
            .map(Main::newPayrollEntry)
            .collect(Collectors.toList());
}

您也可以将相同的代码放在花括号中，但将代码移到静态方法中更易于阅读。

stream.map(employee -> {
    // Long code
    return payroll;
})

如何仅从所需的类（流）获取数据？

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如何仅从所需的 类 （流）获取数据？

How to get data only from desired classes (stream)?

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如何仅从所需的类（流）获取数据？