可以使用窗口函数：

select *
from (
    select t.*, lead(date) over(order by series) lead_date
    from mytable t
) t
where date > lead_date

或者：

select *
from (
    select t.*, lead(series) over(order by date) lead_series
    from mytable t
) t
where series > lead_series

你可以用 lag() :

select t.*
from (select t.*,
             lag(id) over (order by series) as prev_id_series,
             lag(id) over (order by date) as prev_id_date
      from t
     ) t
where prev_id_series <> prev_id_date;

您可以使用如下自连接（假设您的表名为“series”）获取有问题的行及其相应的冲突行：

SELECT s1.id AS row_id, s1.series AS row_series, s1.date AS row_date, 
       s2.id AS conflict_id, s2.series AS conflict_series, s2.date AS conflict_date
FROM series AS s1
JOIN series AS s2
  ON s1.series > s2.series AND s1.date < s2.date;