Oracle SQL。選擇在一天的所有小時全年

Question

我試圖創建一個返回以下結果集的查詢中的每一天（截斷可讀性）：

+---------------+-----------+--------+--------+----------+ 
| DATE_HAPPENED | Twelve_AM | One_AM | Two_AM | Three_AM | (and so on, until 24 hours) 
+---------------+-----------+--------+--------+----------+ 
| 2015-10-01 | 110  | 34 | 92 | 45 | 
+---------------+-----------+--------+--------+----------+ 

這是我使用的代碼（而我不知道它是做）的最佳方式：

SELECT to_char(potty_use_date, 'yyyy-mm-dd HH12') as date_happened, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '12' THEN 1 ELSE 0 END) as Twelve_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '1' THEN 1 ELSE 0 END) as One_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '2' THEN 1 ELSE 0 END) as Two_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '3' THEN 1 ELSE 0 END) as Three_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '4' THEN 1 ELSE 0 END) as Four_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '5' THEN 1 ELSE 0 END) as Five_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '6' THEN 1 ELSE 0 END) as Six_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '7' THEN 1 ELSE 0 END) as Seven_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '8' THEN 1 ELSE 0 END) as Eight_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '9' THEN 1 ELSE 0 END) as Nine_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '10' THEN 1 ELSE 0 END) as Ten_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 AM') = '11' THEN 1 ELSE 0 END) as Eleven_AM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '1' THEN 1 ELSE 0 END) as One_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '2' THEN 1 ELSE 0 END) as Two_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '3' THEN 1 ELSE 0 END) as Three_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '4' THEN 1 ELSE 0 END) as Four_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '5' THEN 1 ELSE 0 END) as Five_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '6' THEN 1 ELSE 0 END) as Six_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '7' THEN 1 ELSE 0 END) as Seven_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '8' THEN 1 ELSE 0 END) as Eight_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '9' THEN 1 ELSE 0 END) as Nine_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '10' THEN 1 ELSE 0 END) as Ten_PM, 
    COUNT(CASE WHEN to_char(potty_use_date, 'HH12 PM') = '11' THEN 1 ELSE 0 END) as Eleven_PM, 
    CASE WHEN to_char(potty_use_date, 'HH12 PM') = '12' THEN 1 ELSE 0 END) as Twelve_PM 
FROM core.potty_usage_statistics 
GROUP BY to_char(potty_use_date, 'yyyy-mm-dd HH12') 
ORDER BY date_happened ASC;   

不過，我得到以下結果：

2015-04-20 08 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 1197 

他們都跨日重複e結果窗口。顯然，我沒有這樣做。我該如何做對？

Answer 1

COUNT()計算非NULL值的數量。如果值爲0或1，則全部爲非NULL。所以，計數與COUNT(*)相同。

兩個選項：

• 更改COUNT()到SUM()
• 取出ELSE 0

就個人而言，我更喜歡第一種方法;但要麼是可行的。

我也應該注意到，您可以使用EXTRACT()用於此目的：

SUM(CASE WHEN extract(hour from potty_use_date) = 0 THEN 1 ELSE 0 END) as Midnight, 

有些人可能會覺得這更容易閱讀。

我還懷疑你想要的GROUP BY和SELECT是：

SELECT to_char(potty_use_date, 'yyyy-mm-dd') as date_happened 
. . . 
GROUP BY to_char(potty_use_date, 'yyyy-mm-dd') 

如果包含小時組件，那麼你將得到一個對角值，24行的每一天，一個小時列爲每個人填充。

Answer 2

也許你可以透視：

select * from 
(
    select 
    trunc(p.potty_use_date) as potty_use_date, 
    to_char(p.potty_use_date, 'HH24') as putty_use_hour, 
    from 
    core.potty_usage_statistics p 
) 
pivot (
    count(putty_use_hour) 
    for putty_use_hour in (
     '00' as "Midnight", '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', 
     '12' as "Noon", '13', '14', '15', '16', '17', '18' as "Dinner time", '19', '20', '21', '22', '23', '24' as "Or is this midnight") 
) 

我沒有手頭有甲骨文，所以我無法測試它，但這應該工作。

有關甲骨文旋轉的更多信息，以及爲什麼你需要一個小時的完整列表，請閱讀SQL Operations: Pivot and Unpivot