Question

我有这个代码

select no1, visit, DATE1, TIME1, STATUS ,hour(diff) * 60 + minute(diff) as diff from( 
select no1, visit, DATE1,

if(
@active or STATUS in ('Pending Active') and STATUS not in ('Stop Clock') ,
timediff(addtime(DATE1,TIME1),@prevTime),0) as diff,

time(@prevTime := addtime(DATE1,TIME1)) as time1,
@active := STATUS in ('Pending Active') and STATUS not in ('Stop Clock'),STATUS from detail,

(select @prevTime :=0,@active :=false) as init where NO_LOG = '03/12/008' and visit = 'On Site Visit' order by no1) as final

，结果如下：

+------+--------------+------------+---------+-------------------+-----+
| no1  |    visit     |    DATE1   |  TIME1  |   STATUS          | diff|
+------+--------------+------------+---------+-------------------+-----+
|74030 |On Site Visit | 2012-03-12 |19:23:00 | Pending           |  0  |
|74031 |On Site Visit | 2012-03-12 |19:44:00 | Pending           |  0  |
|74032 |On Site Visit | 2012-03-12 |20:40:00 | Pending Active    |  56 |
|74033 |On Site Visit | 2012-03-12 |20:45:00 | Pending Active    |  5  |
|74034 |On Site Visit | 2012-03-12 |20:50:00 | Pending Active    |  5  |
|74035 |On Site Visit | 2012-03-12 |20:54:00 | Active            |  4  |
|74036 |On Site Visit | 2012-03-12 |21:30:00 | Close             |  36 |

但是我想要的是，如果访问=现场访问和STATUS =待定活动（第一个待定活动），则第一个待定活动以下的所有deff将为'0'（零）...如下例所示

+------+--------------+------------+---------+-------------------+-----+
| no1  |    visit     |    DATE1   |  TIME1  |   STATUS          | diff|
+------+--------------+------------+---------+-------------------+-----+
|74030 |On Site Visit | 2012-03-12 |19:23:00 | Pending           |  0  |
|74031 |On Site Visit | 2012-03-12 |19:44:00 | Pending           |  0  |
|74032 |On Site Visit | 2012-03-12 |20:40:00 | Pending Active    |  56 |
|74033 |On Site Visit | 2012-03-12 |20:45:00 | Pending Active    |  0  |
|74034 |On Site Visit | 2012-03-12 |20:50:00 | Pending Active    |  0  |
|74035 |On Site Visit | 2012-03-12 |20:54:00 | Active            |  0  |
|74036 |On Site Visit | 2012-03-12 |21:30:00 | Close             |  0  |

也许有人可以帮助我..谢谢

Answer 1

我目前无法提出查询，但我的解决方案是使用UNION ALL并使用WHERE Status =过滤并让第二（或第三）UNION组拥有diff上的值为0。

<强>代码：

SELECT *
FROM (  SELECT no1, visit, DATE1, TIME1, STATUS, 0 AS diff
        FROM detail
        WHERE status IN ('Pending') AND NO_LOG = '03/12/008' AND visit = 'On Site Visit'
        UNION ALL
        SELECT no1, visit, DATE1, TIME1, STATUS, # do the diff calculation here
        FROM detail
        WHERE status IN ('Pending Active') AND NO_LOG = '03/12/008' AND visit = 'On Site Visit'
        LIMIT 1
        UNION ALL
        SELECT no1, visit, DATE1, TIME1, STATUS, 0
        FROM detail
        WHERE status IN ('Pending Active', 'Active', 'Close') AND NO_LOG = '03/12/008' AND visit = 'On Site Visit'
        LIMIT 1, 9999) AS h
ORDER BY no1 ASC

Info about UNION ALL

符合标准后停止计数timediff

1 个答案: