SQL Case 语句子选择内的子选择分组答案

【问题标题】：SQL Case statement subselect within a grouped by subselectSQL Case 语句子选择内的子选择分组
【发布时间】：2013-03-26 16:44:35
【问题描述】：

我不知道如何修改我的查询以使用 information.infoDate 和 otherDates.auxDate 之间的最新日期。

SELECT state, COUNT(type) as EQUIPMENT, 
       (SELECT COUNT(type) 
        FROM information a 
        WHERE a.state = b.state 
              AND storedprocedures.to_date(infoDate) < (SELECT current date - 5 days 
                                                        FROM sysibm.sysdummy1) 
        GROUP BY state) as IDLE,
        ROUND((SELECT COUNT(type) 
               FROM information a 
               WHERE a.state = b.state 
                     AND storedprocedures.to_date(infoDate) < (SELECT current date - 5 days 
                                                               FROM sysibm.sysdummy1)
               GROUP BY state) / CAST(COUNT(type) as float) * 100, 1) as percent
FROM information b
JOIN validStates 
  ON state = vstate
LEFT JOIN otherDates 
  ON typeid = auxtypeid
WHERE state <> '***'
GROUP BY state
ORDER BY state

我已经尝试过使用以下语句而不是 infoDate 的各种方式，但我无法运行查询。有没有更好的方法来使用两个日期中的最新日期？

SELECT CASE WHEN auxDate > infoDate 
            THEN auxDate 
            ELSE infoDate END AS activityDate 
FROM information 
LEFT JOIN otherDates 
       ON typeid = auxtypeid

【问题讨论】：

我们能否获得一些示例起始数据/所需的结果输出？ DB2 的哪个版本/平台？无论您的目标格式是什么（为什么不存储在日期/时间/时间戳类型中），是否有一个“自反”from_date 函数？

标签： sql db2 case sql-subselect

【解决方案1】：

您的 CASE 语句是选择 2 个字段的最新值的最佳方式。为什么查询没有运行？

添加 CASE 语句后...确保您也将 CASE 语句添加到查询的 GROUP BY 部分。

哦，...在您头疼之前 - 您可能需要考虑不使用保留关键字（例如状态、类型等）创建字段。

编辑：哎呀我没有看到 DB2 的标签

【讨论】：

因此，如果您要在 infoDate 所在的位置插入 case 语句，您将如何按 case 语句进行分组？我无法按活动日期分组。
如果您可以使用失败的案例语句回复 SQL，也许我可以提供帮助。起初我没有看到 DB2。在 MS SQL 中，您需要按 CASE 语句进行分组……在 DB2 中，也许您可以简单地 GROUP BY 语句的别名……例如 GROUP BY activityDate

【解决方案2】：

SELECT state,
COUNT (type) as EQUIPMENT, 
(SELECT COUNT(type) FROM information a LEFT JOIN otherDates on typeid=auxtypeid WHERE a.state=b.state AND (CASE WHEN otherDate > infoDate THEN storedprocedures.dec_to_date(otherDate) ELSE storedprocedures.dec_to_date(infoDate) END) <= (select current date - 5 days from sysibm.sysdummy1) GROUP BY state) as IDLE,
ROUND((SELECT COUNT(type) FROM information a WHERE a.state=b.state AND (CASE WHEN otherDate > infoDate THEN storedprocedures.dec_to_date(otherDate) ELSE storedprocedures.dec_to_date(infoDate) END) < (select current date - 5 days from sysibm.sysdummy1) GROUP BY state)/ CAST(COUNT(type) as float)*100,1) as percent
FROM information b
JOIN validStates ON state=vstate
LEFT JOIN otherDates ON typeid=auxtypeid
WHERE state<>'***'
GROUP BY state
ORDER BY state;

【讨论】：

【解决方案3】：

您可能会更好地执行以下操作（直到我对您的表结构/数据了解更多）：

WITH Raw as (SELECT a.state, 
                    MAX(a.infoDate, COALESCE(c.auxDate, '0001-01-01')) as activityDate
             FROM information a
             JOIN validStates b
               ON b.vstate = a.state
             LEFT JOIN otherDates c
                    ON typeId = auxtypeid
             WHERE a.state <> '***'),
     Total as (SELECT state, COUNT(*) as count
               FROM Raw
               GROUP BY state),
     Idle as (SELECT state, COUNT(*) as count
              FROM Raw
              WHERE storedprocedures.to_date(activityDate) < (CURRENT_DATE - 5 DAYS)
              GROUP BY state)
SELECT a.state, a.count as EQUIPMENT, b.count as IDLE, 
       ROUND((100.0 * b.count) / a.count, 1) as percent
FROM Total a
LEFT JOIN Idle b
       ON b.state = a.state
ORDER BY a.state

...它也可能会运行得更快。

请注意，由于使用了LEFT JOINs，如果给定的state 没有“空闲”类型，IDLE 和percent 可能为空。

【讨论】：