如何獲得一天從一個DateTime

Question

假設我有以下Oracle表，我想the last (aka max) entry by aGroup and day where the timepart of dateTime in less than 00:00:15（每日應提取爲dateTime日期部分）

什麼是它的SQL查詢（我不知道如何從dateTime中提取日期）？

   dateTime aGroup 
2011-01-01 00:00:12  a 
2011-01-01 00:00:31  a 
2012-01-01 00:00:09  a 
2012-01-01 00:00:03  a 
2011-01-01 00:00:06  b 
2011-01-01 00:00:14  b 
2011-01-01 00:00:16  b 
2012-01-01 00:00:16  b 
2012-01-01 00:00:11  b 

這是想要我希望結果

   dateTime aGroup 
2011-01-01 00:00:12  a 
2012-01-01 00:00:09  a 
2011-01-01 00:00:14  b 
2012-01-01 00:00:11  b 

Answer 1

這是使用TO_CHAR

SELECT AGROUP, MAX(DATETIME) FROM MYTABLE 
WHERE TO_CHAR(DATETIME,'SS') <'15' 
GROUP BY TO_CHAR(DATETIME,'YYYY-MM-DD'), AGROUP 
ORDER BY AGROUP 

Answer 2

考慮類似這樣的查詢：

SELECT 
    MAX(DateTime), 
    aGroup 
FROM 
    tablename 
WHERE 
    TIME_TO_SEC(TIME(DateTime)) < 15 
GROUP BY 
    DATE(DateTime), 
    aGroup 

這會給你最大的時間（小於00:00 ：15）每個aGroup的每一天。

編輯：更新使用TIME_TO_SEC爲Oracle

Answer 3

的TRUNC function是非常方便的提取天：

SELECT dateTime, aGroup 
FROM 
(
    SELECT dateTime 
      , aGroup 
      , RANK() OVER (PARTITION BY aGroup, TRUNC(dateTime, 'DDD') ORDER BY dateTime DESC) as Rank 
    FROM table 
    WHERE (dateTime - TRUNC(dateTime, 'DDD')) * 24 * 3600 < 15 
) 
WHERE Rank = 1 

線

WHERE (dateTime - TRUNC(dateTime, 'DDD')) * 24 * 3600 < 15 

計算差異天之間的當前日期和一天的開始。
乘以24和3600有多少秒，和過濾如果低於15

Answer 4

select dateTime, agroup 
    from (select dateTime, agroup, 
       row_number() over (partition by agroup, trunc(datetime) 
            order by datetime desc) rn 
      from yourtab 
     where to_number(to_char(datetime, 'hh24miss')) < 15) 
where rn = 1 

會做我的嘗試。請參閱http://sqlfiddle.com/#!4/b4eb98/1