0

我有一个带有日期列的表，我用它来查询以从中获取一组行。

我不存储任何时间值只是日期和 T:00:00:00

我需要锻炼我有数据的未来几天和前几天。

我希望能够执行单个查询和日期以使用下一个和后退 UI 导航和查询来选择当前数据集

所以从这个开始获取最后一组数据

SELECT * 
FROM  `newco_bh_price_data` 
WHERE  `manager_id` =960
AND  `currency_id` =0
AND  `class_id` =  'A'
AND  `price_date` 
IN (SELECT MAX(  `price_date` ) 
FROM  `newco_bh_price_data`
)

我现在需要知道下一个日期是什么，以便查询

SELECT * 
FROM  `newco_bh_price_data` 
WHERE  `manager_id` =960
AND  `currency_id` =0
AND  `class_id` =  'A'
AND  `price_date` = '2002-09-19'

感谢您的帮助和评论

这是数据示例

id  price_date  manager_id  currency_id class_id    value   v_change
554 2002-09-19  960 0   A   8.631   0
553 2002-09-19  960 0   A   9.5635  0
552 2002-09-18  960 0   A   9.1068  0
551 2002-09-18  960 0   A   8.9351  0
550 2002-09-18  960 0   A   9.61    0
549 2002-09-17  960 0   A   9.1868  0
548 2002-09-17  960 0   A   9.0201  0
547 2002-09-17  960 0   A   9.6425  0
546 2002-09-16  960 0   A   9.3377  0
545 2002-09-16  960 0   A   9.3317  0
544 2002-09-16  960 0   A   9.6902  0
543 2002-09-13  960 0   A   9.418   0
542 2002-09-13  960 0   A   9.3669  0
541 2002-09-13  960 0   A   9.7125  0
540 2002-09-12  960 0   A   9.4463  0
539 2002-09-12  960 0   A   9.4383  0
538 2002-09-12  960 0   A   9.7474  0
537 2002-09-11  960 0   A   9.5303  0
536 2002-09-11  960 0   A   9.6592  0
535 2002-09-11  960 0   A   9.8041  0
534 2002-09-10  960 0   A   9.4666  0
533 2002-09-10  960 0   A   9.5878  0
532 2002-09-10  960 0   A   9.7651  0
531 2002-09-09  960 0   A   9.3104  0
530 2002-09-09  960 0   A   9.4841  0
529 2002-09-09  960 0   A   9.7498  0
528 2002-09-06  960 0   A   9.2462  0
527 2002-09-06  960 0   A   9.3823  0
526 2002-09-06  960 0   A   9.7501  0
525 2002-09-05  960 0   A   9.0465  0
4

2 回答 2

0

您真正想要的功能是lag()和lead()。但是，唉，这些在 MySQL 中不可用。我认为实现它们的最简单方法是使用相关子查询。

这是一个例子：

SELECT *,
       (select price_date
        from newco_bh_price_data pd1
        where pd1.manager_id = pd.manager_id and
              pd1.currency_id = pd.currency_id and
              pd1.class_id = pd.class_id and
              pd1.price_date > pd.price_date
        order by price_date
        limit 1
       ) as next_price_date,
       (select price_date
        from newco_bh_price_data pd1
        where pd1.manager_id = pd.manager_id and
              pd1.currency_id = pd.currency_id and
              pd1.class_id = pd.class_id and
              pd1.price_date < pd.price_date
        order by price_date desc
        limit 1
       ) as prev_price_date
FROM  `newco_bh_price_data` pd
WHERE  `manager_id` =960
AND  `currency_id` =0
AND  `class_id` =  'A'
AND  `price_date` = '2002-09-19';

这假定您在查找下一个日期和上一个日期时希望经理、货币和类别相同。

于 2013-05-11T19:44:07.157 回答
0

询问：

SQLFIDDLE示例

SELECT t1.*,
       (SELECT MIN(t2.`price_date`)
        FROM `newco_bh_price_data` t2
        WHERE t2.`price_date`> t1.`price_date`
        AND t2.`manager_id`= t1.`manager_id`
        AND t2.`currency_id` = t1.`currency_id`
        AND t2.`class_id` = t1.`class_id`
        ) AS NextDate,
       (SELECT MAX(t2.`price_date`)
        FROM `newco_bh_price_data` t2
        WHERE t2.`price_date`< t1.`price_date`
        AND t2.`manager_id`= t1.`manager_id`
        AND t2.`currency_id` = t1.`currency_id`
        AND t2.`class_id` = t1.`class_id`
        ) AS PreviousDate
FROM  `newco_bh_price_data` t1
WHERE  t1.`manager_id` =960
AND  t1.`currency_id` =0
AND  t1.`class_id` =  'A'

结果：

|  ID |                       PRICE_DATE | MANAGER_ID | CURRENCY_ID | CLASS_ID | VALUE | V_CHANGE |                         NEXTDATE |                     PREVIOUSDATE |
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------
| 554 | September, 19 2002 00:00:00+0000 |        960 |           0 |        A |     9 |        0 |                           (null) | September, 18 2002 00:00:00+0000 |
| 553 | September, 19 2002 00:00:00+0000 |        960 |           0 |        A |    10 |        0 |                           (null) | September, 18 2002 00:00:00+0000 |
| 552 | September, 18 2002 00:00:00+0000 |        960 |           0 |        A |     9 |        0 | September, 19 2002 00:00:00+0000 | September, 17 2002 00:00:00+0000 |
| 551 | September, 18 2002 00:00:00+0000 |        960 |           0 |        A |     9 |        0 | September, 19 2002 00:00:00+0000 | September, 17 2002 00:00:00+0000 |
| 550 | September, 18 2002 00:00:00+0000 |        960 |           0 |        A |    10 |        0 | September, 19 2002 00:00:00+0000 | September, 17 2002 00:00:00+0000 |
| 549 | September, 17 2002 00:00:00+0000 |        960 |           0 |        A |     9 |        0 | September, 18 2002 00:00:00+0000 | September, 16 2002 00:00:00+0000 |
| 548 | September, 17 2002 00:00:00+0000 |        960 |           0 |        A |     9 |        0 | September, 18 2002 00:00:00+0000 | September, 16 2002 00:00:00+0000 |
于 2013-05-11T19:51:03.007 回答