Question

我有一个带有以下签名的Employee和EmployeeJob类：

public class Employee {

    public Integer employeeId;

    public Integer getEmployeeId() {
        return employeeId;
    }

    public void setEmployeeId(Integer employeeId) {
        this.employeeId = employeeId;
    }
}

public class EmployeeJob {

    private Integer employeeJobId;
    private Employee employee;
    private String jobName;
    public Integer getEmployeeJobId() {
        return employeeJobId;
    }
    public void setEmployeeJobId(Integer employeeJobId) {
        this.employeeJobId = employeeJobId;
    }
    public Employee getEmployee() {
        return employee;
    }
    public void setEmployee(Employee employee) {
        this.employee = employee;
    }
    public String getJobName() {
        return jobName;
    }
    public void setJobName(String jobName) {
        this.jobName = jobName;
    }
}

我想要一个Map<Integer, List<EmployeeJob>>中的一个List<EmployeeJob>，其中键是 Employee 类中的 employeeId 。我如何使用Java 8流按 employee-> employeeId 对该列表进行分组？？

有可能在单个流中求解，还是应该以某种方式拆分它还是采用经典解决方案？

Answer 1

使用Collectors.groupingBy：

Map<Integer, List<EmployeeJob>> map =
    list.stream()
        .collect(Collectors.groupingBy(ej -> ej.getEmployee().getEmployeeId()));

Answer 2

groupingBy()正是这样做的：

Map<Integer, List<EmployeeJob>> map = jobs.stream().collect(groupingBy(job -> job.getEmployee().getEmployeeId()));

不幸的是，由于必须经过Employee类，因此不能使用方法引用。

Answer 3

public static void main(String[] args) {

List<EmployeeJob> listEmployees= new ArrayList<>();
System.out.println(listEmployees);//previous list

Map<Integer, List<EmployeeJob>> newMap =
    listEmployees.stream()
        .collect(Collectors.groupingBy(employee->employee.getEmployee().getEmployeeId()));
System.out.println(newMap); // new expected map
}

要映射的Java 8组列表

3 个答案: