java8 stream自定义分组求和并排序

public static void main(String[] args) {
List<GroupDetailDTO> list = new ArrayList<>();
GroupDetailDTO dto1 = new GroupDetailDTO();
dto1.setHeadsetId(1);
dto1.setTime("2020-01-03");
dto1.setActConcreteTime("a");
dto1.setPlayed(1);
list.add(dto1); GroupDetailDTO dto2 = new GroupDetailDTO();
dto2.setHeadsetId(1);
dto2.setTime("2020-01-01");
dto2.setActConcreteTime("b");
dto2.setPlayed(1);
list.add(dto2); GroupDetailDTO dto3 = new GroupDetailDTO();
dto3.setHeadsetId(1);
dto3.setTime("2020-01-02");
dto3.setActConcreteTime("c");
dto3.setPlayed(1);
list.add(dto3); GroupDetailDTO dto4 = new GroupDetailDTO();
dto4.setHeadsetId(2);
dto4.setTime("2020-01-01");
dto4.setActConcreteTime("d");
dto4.setPlayed(4);
list.add(dto4); Map<GroupDetailDTO, Integer> collect = list.stream().collect(Collectors.groupingBy(o -> new GroupDetailDTO(o.getHeadsetId(), o.getTime()), TreeMap::new, Collectors.summingInt(o -> o.getPlayed())));
collect.forEach((k, v) -> {
System.out.println(k + "@@@" + v);
});
}


@Data
public class GroupDetailDTO implements Comparable<GroupDetailDTO> {
private Integer headsetId;
private String actConcreteTime;
private String time;
private Integer played; public GroupDetailDTO() {
} public GroupDetailDTO(Integer headsetId, String time) {
this.headsetId = headsetId;
this.time = time;
} @Override
public int compareTo(GroupDetailDTO o) {
String a = this.headsetId + "|" + this.getTime();
String b = o.getHeadsetId() + "|" + o.getTime();
return a.compareTo(b);
// return this.getTime().compareTo(o.gtTime());
}
}


结果：

上述代码实现对对象GroupDetailDTO按设备和时间分组求played的和，并且按时间排序了。当初想着既然是按时间排序那么compareTo方法直接使用

return this.getTime().compareTo(o.getTime());

但是结果不正确

分析：分组求和时如果使用了排序，那么分组时对相同属性的合并将直接根据compareTo返回的结果合并对象，所以如果用上述方法将产生相同的时间直接合并，而不是new GroupDetailDTO(o.getHeadsetId(), o.getTime())里面的两个属性都相同才合并。