Question

我有多个用户，每个用户需要10个问题并留下一些时间戳。我理想的做法是找到第n个用户的完整时间，其中n由时间戳的升序而不是用户ID确定。

这不起作用，但为了让你感受到我的需要，如果可以的话，我将非常高兴：

SELECT <N-TH> MAX(u.time) AS maxTime FROM Users u
Group BY u.userId   ORDER BY maxTime

有什么想法？现在我正在运行

SELECT MAX(u.time) AS maxTime FROM Users u
Group BY u.userId   ORDER BY maxTime

将结果作为数组获取，然后找到第n个条目。

Answer 1

许多可能的方法，但只是袖口...可以使用CTE或子查询。尝试

SELECT MAX([maxTime]) AS [LastEntry] FROM (
    SELECT MAX(u.time) AS maxTime,u.userId 
    FROM Users u 
    GROUP BY u.userId,u.time) t

好的，试试这个......

WITH CTE ([maxTime], [userId ]) AS (
    SELECT MAX([time]) AS [maxTime],
        [userId]
    FROM Users
    GROUP BY [userId])
    SELECT MAX([maxTime]) AS [maxTime]
    FROM CTE;

HTH

戴夫

Answer 2

您是否尝试过使用ROW_NUMBER()功能

SELECT * FROM
(
SELECT ROW_NUMBER() OVER (ORDER BY maxTime) as RowNum, MAX(u.time) AS maxTime FROM Users u
Group BY u.userId
) A
WHERE RowNum = NthNumber
ORDER BY maxTime

其中NthNumber是排名编号。

Answer 3

GROUP BY UserId，并根据ROW_NUMBER应用max(u.time)来创建一个排名伪列，然后您可以对其进行过滤

select min_utime, max_utime, userId from (
select min(u.time) min_utime, max(u.time) max_utime, u.userId,
row_number() over (order by max(u.time) asc) as ranker
from Users u
group by u.userId
) Z  where ranker  = @n

Answer 4

按照最长时间排序前10行：

SELECT TOP 10 userId, MAX(time) AS maxtime
FROM Users
GROUP BY userId
ORDER BY maxtime ASC

然后只需颠倒顺序并占据最上一行：

SELECT TOP 1 *
FROM (
  SELECT TOP 10 userId, MAX(time) AS maxtime
  FROM Users
  GROUP BY userId
  ORDER BY maxtime ASC
) s
ORDER BY maxtime DESC;

获取SQL数据库中的第n条记录，其中表是另一个查询的结果

4 个答案: