如何在select语句中使用不同条件进行多个COUNT

Fel*_*rio 3 sql firebird

我在下面做了三个选择,每个选择都返回一个值。如何仅选择并返回三个值,每个值都有其条件?所有信息都在一张表中。

SELECT COUNT (GRA_HRFIM) as "Conferidos antes 17h" FROM GRADE WHERE GRA_HRFIM < '17:00:00' AND GRA_DATADIGITA = '04/28/2021'

SELECT COUNT (GRA_HORADIGITA) as "Importados antes 17h" FROM GRADE WHERE GRA_HORADIGITA < '17:00:00' AND GRA_DATADIGITA = '04/28/2021'

SELECT COUNT (GRA_HRFIM) as "Total pedidos" FROM GRADE WHERE GRA_DATADIGITA = '04/28/2021'
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Kaz*_*Nur 5

GRA_DATADIGITA = '04/28/2021'所有计数的条件都相同,因此在 where 子句中使用它。现在您只需选择即可获得第三次计数,COUNT (GRA_HRFIM)但对于第一次和第二次计数,您需要使用case when语句来选择条件计数。

SELECT COUNT(case when GRA_HRFIM < '17:00:00' then GRA_HRFIM else null end)
         AS "Conferidos antes 17h",
       COUNT(case when GRA_HORADIGITA < '17:00:00' then GRA_HORADIGITA else null end)
         AS  "Importados antes 17h",
       COUNT(GRA_HRFIM)
         AS "Total pedidos"
  FROM GRADE
 WHERE GRA_DATADIGITA = '04/28/2021';
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您还可以避免在语句中提及 case 的 else 部分,因为默认值为 null。

SELECT COUNT(case when GRA_HRFIM < '17:00:00' then GRA_HRFIM end)
             AS "Conferidos antes 17h",
           COUNT(case when GRA_HORADIGITA < '17:00:00' then GRA_HORADIGITA end)
             AS  "Importados antes 17h",
           COUNT(GRA_HRFIM)
             AS "Total pedidos"
      FROM GRADE
     WHERE GRA_DATADIGITA = '04/28/2021'; 
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  • 您可以省略“else null”,因为这是默认值。 (2认同)
  • 如果进一步缩短查询文本,那么还可以进行一项优化。由于我们通常使用 `count( ...` 而不是 `count(distinct ...` ),所以具体值并不重要,只有 null 才重要。所以像 `count(case when GRA_HRFIM &lt; '17:00:00' then 1 end)` 也可以起到同样的作用:-) (2认同)