Java 流：使用不同 object 類型的多個列表創建 Map

Question

我想使用不同 object 類型的多個列表創建對象的 Map。 我想舉個例子，這樣會更清楚，更容易討論。

class Employee {
Long id;
String name;
// getters, setters, and toString
}

class Department {
Long id;
String name;
// getters, setters, and toString
} 

class EmployeeDepartment {
Long id;
Long employeeId;
Long deptId;
// getters, setters, and toString
}

class Subject {
Long id;
String name;
// getters, setters, and toString
}

class EmployeeSubject {
Long id;
Long employeeId;
Long subjectId;
// getters, setters, and toString
}

我現在有

List<Employee> emplList;
List<EmployeeDepartment> emplDeptList;
List<EmployeeSubject> emplSubList;

我想獲得一個 Map ，我可以使用它來搜索員工：deptId 和 subjectId

就像是

Map<List<Long, Long>, List<Employee>> deptSubToEmpMap;
// representing Map<List<deptId, subId>, List<empl>>

// Or
Map<Long, Map<Long, List<Employee>> deptToSubToEmpMap;
// representing Map<deptId, Map<subId, List<empl>>

或任何其他方便的形式，如果它比上述表示更好。

注意：我是 Java Streams 的新手

Answer 1

您需要創建EmployeeSubjectDepartment來保存員工、部門和主題之間的關系，如下所示，

@Getter@Setter
class EmployeeSubjectDepartment{
    Long empId;
    Long deptId;
    Long subId;

    public EmployeeSubjectDepartment(Long empId, Long deptId, Long subId) {
        this.empId = empId;
        this.deptId = deptId;
        this.subId = subId;
    }
}

然后合並emplDeptList和emplSubList得到emplyees

List<EmployeeSubjectDepartment> employeeSubjectDepartments = emplDeptList.stream()
                .flatMap(dept -> emplSubList.stream()
                        .filter(sub -> dept.employeeId.equals(sub.employeeId))
                        .map(sub -> new EmployeeSubjectDepartment(dept.employeeId, dept.deptId, sub.subjectId)))
                .collect(toList());

Map<Long, Map<Long, List<Long>>> result = employeeSubjectDepartments.stream()
        .collect(groupingBy(EmployeeSubjectDepartment::getDeptId,
                groupingBy(EmployeeSubjectDepartment::getSubId, mapping(e -> e.empId, toList()))));

這將導致Map帶有員工 ID List 。

如果您想要員工List ，請創建EmployeeSubjectDepartment ，如下所示，

 class EmployeeSubjectDepartment{
        Employee emp;
        Long deptId;
        Long subId;
    } 

然后合並emplDeptList和emplSubList從emplList獲取和 map 到emplyees

List<EmployeeSubjectDepartment> employeeSubjectDepartments = emplDeptList.stream()
                .flatMap(dept -> emplSubList.stream()
                        .filter(sub -> dept.employeeId.equals(sub.employeeId))
                        .map(sub -> new EmployeeSubjectDepartment(emplList.stream().filter(e -> e.id == sub.employeeId).findAny().orElse(null), dept.deptId, sub.subjectId)))
                .collect(toList());

Map<Long, Map<Long, List<Employee>>> result = employeeSubjectDepartments.stream()
        .collect(groupingBy(EmployeeSubjectDepartment::getDeptId,
                groupingBy(EmployeeSubjectDepartment::getSubId, mapping(e -> e.emp, toList()))));

這可能不是一個完美的解決方案，但您可以獲得所需的結果。