Oracle SQL如何在连接期间计算列值Occurence和Group BY

Question

我正在处理另一个SQL查询，尝试在进行计数和连接表时对记录集合进行分组。 有关构建和填充表的目标，当前查询和附加脚本，请参见下文。

显示所有检查过书籍而不是DVD的客户。 显示客户名称，总结账凭证和总DVD结账。 按客户名字和姓氏对结果进行排序。

SELECT C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME, COUNT(T.TRANSACTION_ID)
FROM customer C
INNER JOIN library_card LC ON C.CUSTOMER_ID = LC.CUSTOMER_ID
INNER JOIN transaction T ON LC.LIBRARY_CARD_ID = T.LIBRARY_CARD_ID
INNER JOIN physical_item P ON T.PHYSICAL_ITEM_ID = P.PHYSICAL_ITEM_ID
INNER JOIN catalog_item CT ON P.CATALOG_ITEM_ID = CT.CATALOG_ITEM_ID 
GROUP BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME
ORDER BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME;

首先运行： https ： //drive.google.com/open？id = 1PYAZV4KIfZtxP4eQn35zsczySsxDM7ls

第二步： https ： //drive.google.com/open？id = 1pAzWmJqvD3o3n6YJqVUM6TtxDafKGd3f

编辑
在Barbaros先生的帮助下，我提出了以下查询，这是更接近的问题。 但是，此查询不会返回任何DVD结果，这让我相信这是一个加入问题。

SELECT C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME, COUNT(CT1.TYPE) AS BOOK_COUNT, COUNT(CT2.TYPE) AS DVD_COUNT
FROM customer C
INNER JOIN library_card LC ON C.CUSTOMER_ID = LC.CUSTOMER_ID
INNER JOIN transaction T ON LC.LIBRARY_CARD_ID = T.LIBRARY_CARD_ID
INNER JOIN physical_item P ON T.PHYSICAL_ITEM_ID = P.PHYSICAL_ITEM_ID
INNER JOIN catalog_item CT1 ON P.CATALOG_ITEM_ID = CT1.CATALOG_ITEM_ID AND CT1.TYPE = 'BOOK'
LEFT OUTER JOIN catalog_item CT2 ON P.CATALOG_ITEM_ID = CT2.CATALOG_ITEM_ID AND CT2.TYPE = 'DVD'
GROUP BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME, CT1.TYPE, CT2.TYPE
ORDER BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME;

Answer 1

使用“条件聚合”（在聚合函数中使用case表达式）

SELECT
    C.CUSTOMER_FIRSTNAME
  , C.CUSTOMER_LASTNAME
  , COUNT( CASE WHEN CT.TYPE = 'BOOK' THEN T.TRANSACTION_ID END ) books
  , COUNT( CASE WHEN CT.TYPE = 'DVD' THEN T.TRANSACTION_ID END ) dvds
FROM customer C
INNER JOIN library_card LC ON C.CUSTOMER_ID = LC.CUSTOMER_ID
INNER JOIN transaction T ON LC.LIBRARY_CARD_ID = T.LIBRARY_CARD_ID
INNER JOIN physical_item P ON T.PHYSICAL_ITEM_ID = P.PHYSICAL_ITEM_ID
INNER JOIN catalog_item CT ON P.CATALOG_ITEM_ID = CT.CATALOG_ITEM_ID
GROUP BY
    C.CUSTOMER_FIRSTNAME
  , C.CUSTOMER_LASTNAME
HAVING
     COUNT( CASE WHEN CT.TYPE = 'BOOK' THEN T.TRANSACTION_ID END ) 
   > COUNT( CASE WHEN CT.TYPE = 'DVD' THEN T.TRANSACTION_ID END )
ORDER BY
    C.CUSTOMER_FIRSTNAME
  , C.CUSTOMER_LASTNAME
;

Answer 2

您可以使用catalog_item表两次（将其视为book和dvds的单独表），并通过HAVING子句进行比较：

SELECT C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME, 
       COUNT(CT1.CATALOG_ITEM_ID) as "Book Checkout", 
       COUNT(CT2.CATALOG_ITEM_ID) as "DVD Checkout"
  FROM customer C
 INNER JOIN library_card LC ON C.CUSTOMER_ID = LC.CUSTOMER_ID
 INNER JOIN transaction T ON LC.LIBRARY_CARD_ID = T.LIBRARY_CARD_ID
 INNER JOIN physical_item P ON T.PHYSICAL_ITEM_ID = P.PHYSICAL_ITEM_ID
  LEFT JOIN catalog_item CT1 ON P.CATALOG_ITEM_ID = CT1.CATALOG_ITEM_ID AND CT1.TYPE = 'BOOK'
  LEFT JOIN catalog_item CT2 ON P.CATALOG_ITEM_ID = CT2.CATALOG_ITEM_ID AND CT1.TYPE = 'DVD'
 GROUP BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME
HAVING COUNT(CT1.CATALOG_ITEM_ID) > COUNT(CT2.CATALOG_ITEM_ID)
 ORDER BY C.CUSTOMER_FIRSTNAME, C.CUSTOMER_LASTNAME;

CUSTOMER_FIRSTNAME  CUSTOMER_LASTNAME   Book Checkout  DVD Checkout
------------------  -----------------   -------------  -------------
Deena               Pilgrim                   3              1
Emile               Cross                     5              2

Answer 3

请尝试在group by子句中删除CT1.TYPE，CT2.TYPE 。