ORACLE - MAX和SUM

Question

下面的查詢將返回按總量排序的最流行的戲劇和行類型組合的列表： 因此，例如：

NAME  ROWTYPE TOTALAMOUNT 
theatre1 middle 200 
theatre2 front 190 
theatre1 front 150 
theatre2 middle 100 

而我需要的僅僅是每家影院的最大：

theatre1 middle 200 
theatre2 front 190 

查詢：

SELECT name, rowtype, sum 
from (select 
name, rowtype, sum(totalamount) sum from trow, fact, theatre 

Where trow.trowid = fact.trowid 
AND 
theatre.theatreid = fact.theatreid 

GROUP BY rowtype, name 
) 

ORDER BY sum DESC, name, rowtype ; 

Answer 1

您可以使用窗口功能如下：

select name, rowtype, sum 
from (select name, rowtype, sum(totalamount) as sumta, 
      max(sum(totalamount)) over (partition by name) as maxsumta 
     from trow join 
      fact 
      on trow.trowid = fact.trowid join 
      theatre 
      on theatre.theatreid = fact.theatreid 
    group byrowtype, name 
    ) nr 
where sumta = maxsumta; 

此外，你應該學會使用正確的，明確的JOIN語法。一個簡單的規則：從不在FROM子句中使用逗號。 始終使用使用正確的顯式JOIN語法。

Answer 2

把你當前的查詢放入一個公共表格表達式中，然後使用窗口函數查找每個影院的最大總量。

WITH cte AS 
(
    SELECT name, rowtype, SUM(totalamount) sum 
    FROM trow 
    INNER JOIN fact 
     ON trow.trowid = fact.trowid 
    INNER JOIN theatre 
     ON theatre.theatreid = fact.theatreid 
    GROUP BY name, rowtype 
) 

SELECT name, rowtype, sum 
FROM 
(
    SELECT name, 
      rowtype, 
      sum, 
      MAX(sum) OVER (PARTITION BY name) maxSum 
    FROM cte 
) t 
WHERE t.sum = t.maxSum