从日期时间结果集中获取最大日期时间

Question

我有以下查询，

SELECT  startTime, lat, lon
FROM table
WHERE category = 12 
AND playerId = 10 
AND CAST(startTime as date) >= (select max(CAST(startTime as date)) from table) 

逻辑是最后（最大）日期取经度，经度。所以，今天所有的一切。问题是我得到了一组结果，因为最大日期是

10-7-2019 00:00:00 52.521 113.658
10-7-2019 00:10:00 42.521 103.658 
10-7-2019 00:20:00 22.521 93.658 
etc...

（（由于同一天的时间不同，我在相同的最大日期获得了很多结果。我没有得到仅一行）

如何从以上设置中获取最大值？我必须做类似

的事情

AND CAST(startTime as time) >= (select max(CAST(startTime as time)) from table) 

在结果集上获得10-7-2019 23:59:59 22.521 93.658。我不知道该怎么做。有点像

SELECT  time, lat, lon
FROM table
WHERE category = 12 
AND playerId = 10 
AND CAST(startTime as date) AS ABC >= (select max(CAST(startTime as date)) from table) 
AND CAST(startTime as time) >= (select max(CAST(startTime as time)) from ABC)

谢谢

Answer 1

您可以只使用order by和limit吗？

SELECT startTime, lat, lon
FROM table
WHERE category = 12 AND playerId = 10 
ORDER BY startTime DESC
LIMIT 1;

如果只想在数据的最后一个日期得到结果（整体），则可以使用子查询或窗口函数：

SELECT startTime, lat, lon
FROM (SELECT t.*,
             DENSE_RANK() OVER (PARTITION BY date(startTime) DESC) as seqnum
      FROM table t
     ) t
WHERE category = 12 AND playerId = 10 AND
      seqnum = 1
ORDER BY startTime DESC
LIMIT 1;

Answer 2

您能试试这个吗..

SELECT  startTime, lat, lon
FROM table
WHERE category = 12 
  AND playerId = 10 
  AND CAST(startTime as datetime) >= (select max(CAST(startTime as datetime)) from table)

例如。。

SELECT  Max(Cast(CreatedOn as datetime))
from TimeSheetJob

结果：-

2019-10-04 15:52:29.080