问题描述
我有以下要求，我有 2 个列表，一个是 EmployeeList 和 DepartmentList 现在我想根据下面示例中的字段将这 2 个列表合并/附加到另一个 List(finalList) 中。
List<Employee> list = List.of( new Employee(1,"aba",101),new Employee(2,"cdc",102),new Employee(3,"ded",102));
List<Department> list1 = List.of( new Department(101,"Dep1"),new Department(102,"Dep2"));
List <EmployeeDepartment> finalList = List.empty();
以下是类：Employee、Department 和 EmployeeDepartment。
public class EmployeeDepartment {
Integer deptno;
Integer empid;
String empname;
---setters and Getters
}
public class Department {
Integer deptno;
String depname;
---setters and Getters
}
public class Employee {
Integer empid;
String empname;
Integer deptno;
---setters and Getters
}
解决方法
您可以对两个列表进行迭代并添加新的 EmployeeDepartment 对象
List<EmployeeDepartment> finalList = new ArrayList<>();
for (Employee e : list) {
for (Department d : list1) {
finalList.add(new EmployeeDepartment(e.getEmpid(),e.getEmpname(),d.getDeptno()));
}
}
或者使用 Stream
List<EmployeeDepartment> finalList = list.stream()
.flatMap(e -> list1.stream().map(d -> new EmployeeDepartment(e.getEmpid(),d.getDeptno())))
.collect(Collectors.toList());
我不确定预期的输出是什么（还有 EmployeeDepartment 具有 Employee 具有的所有字段），但是如果您需要在两个列表中包含元素的所有组合，那么 {{ 1}} 正是您所需要的：
crossProduct你得到的是一个带有 6 个元组的迭代器。
如果您需要使用 class Lol {
public static void main(String[] args) {
List<Employee> list = List.of(
new Employee(1,"aba",101),new Employee(2,"cdc",102),new Employee(3,"ded",102)
);
List<Department> list1 = List.of(
new Department(101,"Dep1"),new Department(102,"Dep2")
);
final Iterator<Tuple2<Employee,Department>> tuple2s = list.crossProduct(list1);
System.out.println(tuple2s.size());
}
}
为每个 Department 获取 Employee，那么最好将第二个集合转换为 deptno 并使用它来映射第一个合集。