[英]Java 8 stream group by min and max
假设您对一个employees表运行一个SQL查询:
SELECT department, team, MIN(salary), MAX(salary)
FROM employees
GROUP BY department, team
然后在Java客户端中,通过进行如下所示的DAO调用,将结果集映射到Aggregate
实例列表:
List<Aggregate> deptTeamAggregates = employeeDao.getMinMaxSalariesByDeptAndTeam()
并且“集合”具有部门,团队,minSalary,maxSalary的吸气方法,并且有一个Pair<T, T>
元组
将结果集映射到下面的两个映射中,最清晰,可能也是最佳的方法是:
Map<String, Pair<Integer, Integer>> byDepartmentMinMax = ...
Map<Pair<String, String>, Pair<Integer, Integer>> byDepartmentAndTeamMinMax = ...
我知道我可以用不同的方式映射结果集和/或两次访问数据库并以更简单的方式实现相同的目的,但是我更多的是了解Java 8功能。
预先感谢您的输入。
class Pair<T, U> {
public final T x;
public final U y;
public Pair(T x, U y) {
this.x = x;
this.y = y;
}
}
Collector<Aggregate, ?, Pair<Integer, Integer>> aggregateSalary =
mapping(a -> new Pair<>(a.getMinSalary(), a.getMaxSalary()),
reducing(new Pair<>(Integer.MAX_VALUE, Integer.MIN_VALUE),
(a, b) -> new Pair<>(Math.min(a.x, b.x), Math.max(a.y, b.y))));
Map<String, Pair<Integer, Integer>> byDepartmentMinMax =
deptTeamAggregates.stream()
.collect(groupingBy(a -> a.getDepartment(), aggregateSalary));
Map<Pair<String, String>, Pair<Integer, Integer>> byDepartmentAndTeamMinMax =
deptTeamAggregates.stream()
.collect(toMap(a -> new Pair<>(a.getDepartment(), a.getTeam()), a -> new Pair<>(a.getMinSalary(), a.getMaxSalary())));
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