Question

我有2个员工列表。一个列表包含姓名和employeeId，第二个列表包含employeeId和手机号码。员工ID是主键。要求是使用流媒体获取包含ID，名称和移动电话号码的列表。

public class MainApp {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Employee emp1 = new Employee(101, "Shiv1");
        Employee emp2 = new Employee(102, "Shiv2");
        Employee emp3 = new Employee(103, "Shiv3");
        Employee emp4 = new Employee(104, "Shiv4");

        Employee emp5 = new Employee(101, 00001);
        Employee emp6 = new Employee(101, 00002);
        Employee emp7 = new Employee(101, 00003);
        Employee emp8 = new Employee(101, 00004);

        List<Employee> employeeNameList = new ArrayList<Employee>();
        employeeNameList.add(emp1);
        employeeNameList.add(emp2);
        employeeNameList.add(emp3);
        employeeNameList.add(emp4);

        List<Employee> employeeMobileList = new ArrayList<Employee>();
        employeeMobileList.add(emp5);
        employeeMobileList.add(emp6);
        employeeMobileList.add(emp7);
        employeeMobileList.add(emp8);

        employeeNameList.stream()
            .filter(item -> item.getId() == 3)
            .map(i -> i.setMobileNo(9089));
    }
}

Answer 1

您可以使用第一个列表创建empId到其name的映射。

Map<Integer, String> empIdToName = employeeNameList.stream()
        .collect(Collectors.toMap(Employee::getId, Employee::getName, (a, b) -> a));

使用这种映射来进一步创建对象，同时遍历第二个并查找该映射，例如：

List<Employee> employees = employeeMobileList.stream()
        .filter(item -> empIdToName.containsKey(item.getId()))
        .map(i -> new Employee(i.getId(), empIdToName.get(i.getId()), i.getMobileNo()))
        .collect(Collectors.toList());

使用两个列表在流中进行流

1 个答案: