Oracle - 如何以最小/最大速率显示[公寓]的总数（嵌套聚合函数问题）

Question

我想发现最简单/性能最差的方法，在一行数据中显示组内有多少记录与正在分组的数据中的最小/最大速率匹配。在我的示例中，建筑物正在分组。

在我提供的伪代码中，以及现有的列我还想提供两个列，显示每个建筑物中有多少公寓的费率与每栋建筑内的最低/最高费率相匹配。直觉上我认为我需要做的就是计算公寓费率与小组的最小/最大费率相匹配的每次出现。这里的问题是我现在正在嵌套聚合函数，但是没有成功。也许我错过了一些简单的东西。

在这个版本的查询（伪代码比我的实际代码简单得多）我可以包装查询然后使用最小/最大值来比较外部查询中的街道速率，但我正在寻找一个简单的方法如果我能够在CASE和SUM（）中嵌套MIN（速率），可能就像我的伪代码一样简单。除了使这个查询变得复杂之外，我不确定一个简单的方法。

不起作用的具体线是LINE 20: SUM(CASE WHEN rate = MIN(rate) THEN 1 ELSE 0 END) AS tot_min_rates。我已经注释掉了。

WITH rates AS
(
  SELECT 1001 AS apt_Key, 5  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1002 AS apt_Key, 4  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1003 AS apt_Key, 17 AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1004 AS apt_Key, 10 AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1005 AS apt_Key, 4  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1006 AS apt_Key, 7  AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1007 AS apt_Key, 5  AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1008 AS apt_Key, 13 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1009 AS apt_Key, 10 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1010 AS apt_Key, 13 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1011 AS apt_Key, 8  AS rate, 130 AS building FROM dual
) /* END rates CTE */

SELECT COUNT(apt_key) AS tot_apts, building
  , ROUND(AVG(rate),1) AS avg_rate
  , MIN(rate) AS min_rate
  , MAX(rate) AS max_rate  
  -- SUM(CASE WHEN rate = MIN(rate) THEN 1 ELSE 0 END) AS tot_min_rates  
FROM rates
GROUP BY building
;

Answer 1

据我所知，似乎没有一种方法可以在不使用子选择或虚拟表或CTE的情况下在同一个简单查询中使用它。尝试执行rate = MAX（rate）只是SQL无法识别。从我可以看出，当你尝试这样做时，没有数据库引擎可以很好地运行。游民。

我发现下面的链接是MS ACCESS，但它是同样的问题。

Use SQL to display count of min and count of max within group

虽然上述链接中的解决方案可行，但我更喜欢易读性和重用公用表表达式（CTE）来组织我的查询，我正在使用的解决方案如下。

WITH rates AS
(
  SELECT 1001 AS apt_Key, 5  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1002 AS apt_Key, 4  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1003 AS apt_Key, 17 AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1004 AS apt_Key, 10 AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1005 AS apt_Key, 4  AS rate, 120 AS building FROM dual UNION ALL
  SELECT 1006 AS apt_Key, 7  AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1007 AS apt_Key, 5  AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1008 AS apt_Key, 13 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1009 AS apt_Key, 10 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1010 AS apt_Key, 13 AS rate, 130 AS building FROM dual UNION ALL
  SELECT 1011 AS apt_Key, 8  AS rate, 130 AS building FROM dual
) /* END rates CTE */

/* Use the below to extract the min/max values per Group aka Building in this case */
, building_min_max AS
(
SELECT building, MIN(rate) AS min_rate, MAX(rate) AS max_rate, ROUND(AVG(rate), 1) AS avg_rate
FROM rates
GROUP BY building
)

/* Using a Detailed view mark each max and min rate so we can sum the up later */
SELECT COUNT(apt_key) AS tot_apts, r.building
  , ROUND(AVG(rate), 1) AS avg_rate
  , MIN(rate) AS min_rate
  , MAX(rate) AS max_rate
  , SUM(CASE WHEN r.rate = mm.min_rate THEN 1 ELSE 0 END) AS tot_min_rates
  , SUM(CASE WHEN r.rate = mm.max_rate THEN 1 ELSE 0 END) AS tot_max_rates
FROM rates r
  JOIN building_min_max mm ON mm.building = r.building
GROUP BY r.building
;