SUM（）GROUP BY列的結果

Question

表格'volcado'：id，reserva，candidad

用以下查詢：

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   ║
╚══════╝

我需要查詢的結果將是：3

如何在同一查詢中對Suma的行進行SUM（）？

我不使用它運行：

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：

子查詢返回1行以上

提前致謝。

Answer 1

試試這個

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，將子查詢用作表而不是列。

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

您可以使用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

您可以根據需要簡化查詢。 因此， having使用case ，不如使用：

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