如何在 Java 中实现 SQL GROUP BY？
先决条件: Java 集合、 Java 数组列表、Java 中的比较器接口
给定一个包含多条记录的列表对象，我们的任务是在 Java 中对列表的字段执行 SQL group by 。
Input: A list containing employee records
in random order.
Output: List sorted first by department
and then by age in each department
进场:
- 对部门和年龄分别使用比较器对列表进行排序。
- 然后我们首先使用 collections.sort() 方法进行排序。
- 我们将使用比较器界面的 【比较() 方法对列表进行排序。使用这种方法非常简单，可以用来链接多个比较器，从而使过程更容易。
下面是上述方法的实现:
程序:
// Java code to implement SQL GROUP BY
import java.util.*;
// Create the employee class
class Employee {
int empid;
String name;
int age;
String dept;
int salary;
public Employee(int empid, String name, int age,
String dept, int sal)
{
this.name = name;
this.empid = empid;
this.age = age;
this.dept = dept;
this.salary = sal;
}
public String getDept()
{
return this.dept;
}
public static void main(String[] args)
{
// create the employee list
List<Employee> empData
= new ArrayList<Employee>();
// add values to list by
// using the add() method
// of the list interface
empData.add(new Employee(1, "Ajay",
25, "Technical", 35000));
empData.add(new Employee(3, "Chandan",
22, "Technical", 30000));
empData.add(new Employee(4, "Arjun",
30, "Management", 54000));
empData.add(new Employee(2, "Arun",
28, "Sales", 9000));
empData.add(new Employee(8, "Anmol",
40, "Sales", 15000));
empData.add(new Employee(9, "Vivek",
20, "Management", 8000));
empData.add(new Employee(10, "Nikhil",
27, "Sales", 7000));
empData.add(new Employee(5, "Rahul",
45, "Management", 60000));
empData.add(new Employee(6, "Ganesh",
32, "Sales", 35000));
empData.add(new Employee(7, "Vishal",
35, "Technical", 40000));
empData.add(new Employee(11, "Anmol",
23, "Sales", 15000));
empData.add(new Employee(12, "Vivek",
29, "Management", 8000));
empData.add(new Employee(13, "Nikhil",
30, "Technical", 7000));
System.out.println("\n Employee list"
+ " before sorting...\n");
System.out.println(" ID Name Age"
+ " Department Salary \n");
for (Employee e : empData) {
System.out.format(" %2d %7s %d %10s %d \n",
e.empid, e.name, e.age, e.dept, e.salary);
}
// Sorting the list by department
// and age by using the thenComparing() method
Collections.sort(empData,
new DepartmentComparator()
.thenComparing(new AgeComparator()));
System.out.println("\n Employee list after sorting...\n");
System.out.println(" ID Name Age Department Salary \n");
for (Employee emp : empData) {
System.out.format(" %2d %7s %d %10s %d \n",
emp.empid, emp.name, emp.age,
emp.dept, emp.salary);
}
}
}
class DepartmentComparator implements Comparator<Employee> {
public int compare(Employee e1, Employee e2)
{
return e1.dept.compareTo(e2.dept);
}
}
class AgeComparator implements Comparator<Employee> {
public int compare(Employee e1, Employee e2)
{
if (e1.age == e2.age)
return 0;
else if (e1.age > e2.age)
return 1;
else
return -1;
}
}
Output:
Employee list before sorting...
ID Name Age Department Salary
1 Ajay 25 Technical 35000
3 Chandan 22 Technical 30000
4 Arjun 30 Management 54000
2 Arun 28 Sales 9000
8 Anmol 40 Sales 15000
9 Vivek 20 Management 8000
10 Nikhil 27 Sales 7000
5 Rahul 45 Management 60000
6 Ganesh 32 Sales 35000
7 Vishal 35 Technical 40000
11 Anmol 23 Sales 15000
12 Vivek 29 Management 8000
13 Nikhil 30 Technical 7000
Employee list after sorting...
ID Name Age Department Salary
9 Vivek 20 Management 8000
12 Vivek 29 Management 8000
4 Arjun 30 Management 54000
5 Rahul 45 Management 60000
11 Anmol 23 Sales 15000
10 Nikhil 27 Sales 7000
2 Arun 28 Sales 9000
6 Ganesh 32 Sales 35000
8 Anmol 40 Sales 15000
3 Chandan 22 Technical 30000
1 Ajay 25 Technical 35000
13 Nikhil 30 Technical 7000
7 Vishal 35 Technical 40000
注:在上述输出中，列表按部门分组，部门进一步按员工年龄排序。
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