如何使用数据透视表进行分组

Question

有一张这样的桌子：，

name  ZK04_COUNT  ZK05_COUNT  ZK04_PRICE  ZK05_PRICE
A     15           0             150.00      0 
A     0            20              0         223.00
B     11           0             75.00       0
B     0            24              0         250.00

我正试图将它转变成这样的：

name  ZK04_COUNT  ZK05_COUNT  ZK04_PRICE  ZK05_PRICE
A         15          20        150.00       223.00
B         11          24         75.00       250.00

我的代码是这样的：

SELECT *
FROM (
      SELECT 
       r.name
      ,r.AUART
      , CASE WHEN r.AUART = 'ZK05' THEN count(r.MATNR) ELSE '' END AS ZK05_COUNT
      , CASE WHEN r.AUART = 'ZK04' THEN count(r.MATNR) ELSE '' END AS ZK04_COUNT
      , CONVERT(money, sum(CAST(netpr as MONEY)), 1) as ToplamTutar  
      FROM GARANTI_YP_RAPORU as r
      group by r.name ,r.AUART
     ) as gTablo
PIVOT
(
  SUM(ToplamTutar)

  FOR gTablo.AUART IN ([ZK05],[ZK04])
)
AS p1

我期望单行输出。怎么能这样？

Answer 1

这真的不需要 PIVOT。
仅按名称对它们进行分组，然后使用条件聚合应该可以。

SELECT r.name
, COUNT(CASE WHEN r.AUART = 'ZK04' THEN r.MATNR END) AS ZK04_COUNT
, COUNT(CASE WHEN r.AUART = 'ZK05' THEN r.MATNR END) AS ZK05_COUNT
, CAST(SUM(CASE WHEN r.AUART = 'ZK04' THEN r.netpr ELSE 0 END) AS MONEY) AS ZK04_PRICE
, CAST(SUM(CASE WHEN r.AUART = 'ZK05' THEN r.netpr ELSE 0 END) AS MONEY) AS ZK05_PRICE
FROM GARANTI_YP_RAPORU as r
GROUP BY r.name
ORDER BY r.name