java - 使用两个列表在流中进行流传输

Question

我有 2 名员工名单。第一个列表包含姓名和员工 ID，第二个列表包含员工 ID 和手机号码。员工 ID 是主键。要求是使用流式传输方式获取包含 id、姓名和手机号码的列表。

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

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));
   

2. 使用这样的映射进一步创建对象，同时迭代第二个映射并查找此映射，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());