java简单实现lambda将数据行转列

1、需求一个级别可以有多个档级，级别和档位可能会变动

2、数据库存储的元数据是级别+档次+薪资

代码实现：

public class Test {

    public static void main(String[] args) {
        List list = initDemoList();
        // 级别分组
        List responseList = list.stream().collect(Collectors.groupingBy(Demo::getLevel)).entrySet().stream()
            // 级别排序
            .sorted(Map.Entry.comparingByKey()).map(integerListEntry -> {
                // 新数组接收
                List newList = new ArrayList();
                newList.add(integerListEntry.getKey());
                // 档位排序
                newList.addAll(integerListEntry.getValue().stream().sorted((o1, o2) -> o1.getGrade() - o2.getGrade())
                    // 收集薪资
                    .map(Demo::getSalary).collect(Collectors.toList()));
                return newList;
            }).collect(Collectors.toList());
        System.out.println(JSONUtil.toJSONString(responseList));
    }

    public static List initDemoList() {
        List list = new ArrayList<>();
        Demo demo1 = new Demo(1, 1, 3, 1003D, "1");
        Demo demo2 = new Demo(2, 1, 1, 1001D, "1");
        Demo demo3 = new Demo(3, 1, 2, 1002D, "1");

        Demo demo4 = new Demo(21, 2, 2, 1002D, "1");
        Demo demo5 = new Demo(22, 2, 1, 1001D, "1");
        Demo demo6 = new Demo(23, 2, 3, 1003D, "1");
        list.add(demo1);
        list.add(demo2);
        list.add(demo3);
        list.add(demo4);
        list.add(demo5);
        list.add(demo6);
        return list;
    }

    @Data
    @AllArgsConstructor
    static class Demo {

        /**
         * 级别或者薪级
         */
        private Integer id;

        /**
         * 级别或者薪级
         */
        private Integer level;

        /**
         * 档次
         */
        private Integer grade;

        /**
         * 级别薪级工资标准
         */
        @ApiModelProperty(value = "级别薪级工资标准")
        private Double salary;

        /**
         * 类别
         */
        private String salaryType;

    }

}

结果：