SUM（）組的結果BY列

Question

表 'volcado'：ID，釀，cantidad

隨着以下查詢：

SELECT (CASE 
     WHEN sum(cantidad) > 0 THEN 1 
     WHEN sum(cantidad) <= 0 THEN 0 
     END) AS suma 
FROM volcado 
GROUP BY reserva 

結果是：

╔══════╗ 
║ suma ║ 
╠══════╣ 
║ 1 ║ 
║ 1 ║ 
║ 0 ║ 
║ 1 ║ 
║ 0 ║ 
╚══════╝ 

我需要查詢的結果是：3

我該如何SUM（）在同一個查詢上的'suma'行？

我不明白這一點與運行：

SELECT SUM(SELECT (CASE 
     WHEN sum(cantidad) > 0 THEN 1 
     WHEN sum(cantidad) <= 0 THEN 0 
     END) AS suma FROM volcado GROUP BY reserva) 
FROM volcado 

錯誤：

提前

Subquery returns more than 1 row

感謝。

Answer 1

嘗試這種方式

SELECT SUM(a.suma) from 
    (SELECT (CASE 
     WHEN sum(cantidad) > 0 THEN 1 
     WHEN sum(cantidad) <= 0 THEN 0 
     END) AS suma FROM volcado GROUP BY reserva)a 

Answer 2

這應該工作，使用子查詢作爲一個表而不是列。

SELECT SUM(suma) from 
    (SELECT (CASE 
     WHEN sum(cantidad) > 0 THEN 1 
     WHEN sum(cantidad) <= 0 THEN 0 
     END) AS suma FROM volcado GROUP BY reserva) 

Answer 3

You can use CTE:

;WITH cte AS 
    (SELECT CASE 
       WHEN sum(cantidad) > 0 THEN 1 
       WHEN sum(cantidad) <= 0 THEN 0 
      END suma 
    FROM volcado 
    GROUP BY reserva) 
SELECT sum(suma) 
FROM cte 

Answer 4

可以簡化查詢，如果你喜歡。因此，而不是使用case使用having：

SELECT COUNT(*) 
FROM (SELECT reserva 
     FROM volcado 
     GROUP BY reserva 
     HAVING SUM(cantidad) > 0 
    ) r;