Question

public class FetchVarableList {

public static void main(String[] args) {

    List<List<Employee>> empsList = new ArrayList<>();

    Employee e1 = new Employee(1, "Abi", "Fin", 2000);
    Employee e2 = new Employee(2, "Chandu", "OPs", 5000);
    Employee e3 = new Employee(3, "mahesh", "HR", 8000);
    Employee e4 = new Employee(4, "Suresh", "Main", 1000);


    List<Employee> empList = new ArrayList<>();
    empList.add(e1); empList.add(e2); 


    List<Employee> empList2 = new ArrayList<>();

    empList2.add(e3); empList2.add(e4);


    empsList.add(empList);

    empsList.add(empList2);
}

}

在此代码上方是empList2中的empList，e3，e4中的雇员e1，e2的列表这两个雇员名单都添加到empslist，我想在单个整数列表中获取所有员工编号

如何从java8中的一个列表中获取所有员工编号列表

Answer 1

List<Integer> numbers = empsList.stream()
                                .flatMap(List::stream)
                                .map(Employee::getNumber)
                                .collect(Collectors.toList());

Answer 2

List<Integer> numbers = Stream.concat(empList.stream(), empList2.stream())
     .map(Employee::getNumber)
     .collect(Collectors.toList());

Answer 3

结合 JB Nizet 和 ahmet kamaran 的答案，您可以避免创建列表列表（empsList ）通过连接2个列表：

List<Integer> numbers = Stream.concat(empList.stream(), empList2.stream())
     .map(Employee::getNumber)
     .collect(Collectors.toList());

从java列表中提取列表值

3 个答案: