Question

我在查询mysql中遇到了问题我想找到2日期除以日期之间的每个日期的小时数例如，

我已经开始 14/12/2014 09:00:00 AM ，并且结束 17/12/2014 03:30:00 AM

我想这样结果

14/12/2014 - 15小时
15/12/1014 - 24小时
16/12/2014 - 24小时
17/12/2014 - 3,5小时

这可能在MySQL或Oracle SQL查询中吗？

Answer 1

使用MySQL TIMESTAMPDIFF或Oracle EXTRACT函数计算Datetime类型（第2和第3个参数）的两个值字段之间的差异。传递的第一个参数是您选择的测量值来计算差异。这可能是HOUR，MINUTE，SECOND，DAY等。

SELECT 
  `startdate` AS `date`, 
  TIMESTAMPDIFF(
    HOUR, 
    `startdate`, 
    `enddate`) AS `hours_delta`
FROM my_table

结果将是：

___ _______日期____ hours_delta

14/12/2014 ________ 66 ________

15/12/1014 ________..._________

Answer 2

最后，我找到了oracle sql的查询...
感谢您的帮助

SELECT
  TO_DATE (START_DATE + COLUMN_VALUE - 1, 'DD/MM/YYYY') AS THE_DATE,
  CASE
    WHEN TO_DATE (START_DATE, 'DD/MM/YYYY') = TO_DATE (END_DATE, 'DD/MM/YYYY')
      THEN (END_DATE - START_DATE)*24
    WHEN (START_DATE + COLUMN_VALUE - 1) = START_DATE
      THEN (23 - TO_CHAR (START_DATE, 'HH24')) + ((60 - TO_CHAR (START_DATE, 'MI'))/60) + ((360 - TO_CHAR (START_DATE, 'SS'))/360)
    WHEN TO_DATE (START_DATE + COLUMN_VALUE - 1, 'DD/MM/YYYY') <> TO_DATE (END_DATE, 'DD/MM/YYYY')
      THEN 24
    ELSE TO_CHAR (END_DATE, 'HH24') + 
         (TO_CHAR (END_DATE, 'MI')/60) + 
         (TO_CHAR (END_DATE, 'SS')/360) END
FROM MY_TABLE,
     TABLE( CAST( MULTISET (
       SELECT
         LEVEL L
       FROM DUAL
       CONNECT BY LEVEL <= END_DATE - START_DATE + 1) AS SYS.ODCINUMBERLIST))

Answer 3

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE MY_TABLE ( ID, START_DATE, END_DATE ) AS
          SELECT 1, TO_DATE( '14/12/2014 09:00:00AM', 'DD/MM/YYYY HH:MI:SSAM'), TO_DATE( '17/12/2014 03:30:00AM', 'DD/MM/YYYY HH:MI:SSAM') FROM DUAL
UNION ALL SELECT 2, TO_DATE( '20/12/2014 09:00:00AM', 'DD/MM/YYYY HH:MI:SSAM'), TO_DATE( '20/12/2014 03:30:00PM', 'DD/MM/YYYY HH:MI:SSAM') FROM DUAL

查询1 ：

WITH Differences AS (
  SELECT ID,
         TRUNC( START_DATE ) AS Day,
         LEAST(
           TRUNC( START_DATE ) + INTERVAL '1' DAY,
           END_DATE
         ) - START_DATE AS Diff
  FROM   MY_TABLE
  UNION
  SELECT ID,
         COLUMN_VALUE,
         CASE WHEN COLUMN_VALUE < TRUNC( END_DATE )
              THEN 1
              ELSE END_DATE - COLUMN_VALUE
              END
  FROM   MY_TABLE,
         TABLE( CAST( MULTISET (
           SELECT TRUNC( START_DATE ) + LEVEL
           FROM   DUAL
           WHERE  TRUNC( START_DATE ) + LEVEL <= TRUNC( END_DATE )
           CONNECT BY
                  TRUNC( START_DATE ) + LEVEL <= TRUNC( END_DATE )
         ) AS SYS.ODCIDATELIST ) )
)
SELECT ID,
       TO_CHAR( Day, 'YYYY-MM-DD' ) AS Day,
       TO_CHAR( TO_NUMBER( Diff * 24 ), '99D90' )
         || ' hour ' AS Hours
FROM   DIFFERENCES

<强> Results ：

| ID |        DAY |        HOURS |
|----|------------|--------------|
|  1 | 2014-12-14 |  15.00 hour  |
|  1 | 2014-12-15 |  24.00 hour  |
|  1 | 2014-12-16 |  24.00 hour  |
|  1 | 2014-12-17 |   3.50 hour  |
|  2 | 2014-12-20 |   6.50 hour  |

在mysql或oracle SQL中的日期之间减少/增加

3 个答案: