当计数不同的结果为0时如何隐藏行？ (Postgres)

Question

我有这个代码

 SELECT "School"."name" AS "School",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'A') AS "A",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'B') AS "B",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'C') AS "C",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'D') AS "D"
FROM "public"."user_tokens"
LEFT JOIN "public"."users" "User" ON "public"."user_tokens"."user_id" = "User"."id"
LEFT JOIN "public"."user_roles" "User_2" ON "public"."user_tokens"."user_id" = "User_2"."user_id"
LEFT JOIN "public"."roles" "Role" ON "User_2"."role_id" = "Role"."id" LEFT JOIN "public"."schools" "School" ON "User_2"."school_id" = "School"."id"
GROUP BY "School"."name"
ORDER BY "B" desc

结果是这样的：

  School         A        B        C         D
--------------------------------------------------
    P            5        6       10         6
    Q            1        0        0         0
    R            2        7        0         6
    S            0        8        9         0

是否可以隐藏包含值“0”的整行？在这种情况下，结果应该只是 School P。

之后，如何区分不包含零值的“学校”？对于这种情况，结果应该是 1

谢谢。

Answer 1

您使用having 子句并重复表达式

having count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'A') > 0 and
       count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'B') > 0 and
       count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'C') > 0 and
       count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'D') 

虽然 Postgres 允许在 GROUP BY 中使用列别名，但它不允许在 HAVING 子句中使用列别名上的表达式（在我看来，这是扩展 SQL 标准的一种奇怪方式）。

现在，您还可以简化和改进您的查询。首先，您需要匹配，所以只需使用inner join。其次，使用表别名。我也摆脱了双引号 - 这是一个非常糟糕的主意：

 SELECT s.name AS School,
        count(distinct u.id) filter (where ut.app_name = 'A') AS A,
        count(distinct u.id) filter (where ut.app_name = 'B') AS B,
        count(distinct u.id) filter (where ut.app_name = 'C') AS C,
        count(distinct u.id) filter (where ut.app_name = 'D') AS D
FROM "public"."user_tokens" ut JOIN
     "public"."users" u
      ON ut.user_id = u.id JOIN
      "public"."user_roles" ur
      ON ut.user_id = ur.user_id JOIN
      "public"."roles" r
      ON ur.role_id = r.id JOIN
      "public"."schools" s
      ON ur.school_id = s.id
GROUP BY s.name
having count(distinct u.id) filter (where ut.app_name = 'A') > 0 and
       count(distinct u.id) filter (where ut.app_name = 'B') > 0 and
       count(distinct u.id) filter (where ut.app_name = 'C') > 0 and
       count(distinct u.id) filter (where ut.app_name = 'D') 
ORDER BY "B" desc

Answer 2

GROUP BY "School"."name"
HAVING count... > 0
      and count... > 0
ORDER BY "B" desc

Answer 3

此外，您可以通过将查询封装为另一个SELECT 的表表达式，使A、B、C 和D 列可供操作。在这个外部SELECT 中，通过添加过滤谓词很容易过滤掉行：

WHERE "A" <> 0 and "B" <> 0 and "C" <> 0 and "D" <> 0

因此，您的查询可能如下所示：

SELECT
  *,
  count(*) over() as total_rows
from (
  SELECT "School"."name" AS "School",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'A') AS "A",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'B') AS "B",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'C') AS "C",
        count(distinct "User"."id") filter (where "public"."user_tokens"."app_name" = 'D') AS "D"
  FROM "public"."user_tokens"
  LEFT JOIN "public"."users" "User" ON "public"."user_tokens"."user_id" = "User"."id"
  LEFT JOIN "public"."user_roles" "User_2" ON "public"."user_tokens"."user_id" = "User_2"."user_id"
  LEFT JOIN "public"."roles" "Role" ON "User_2"."role_id" = "Role"."id" LEFT JOIN "public"."schools" "School" ON "User_2"."school_id" = "School"."id"
  GROUP BY "School"."name"
) x
WHERE "A" <> 0 and "B" <> 0 and "C" <> 0 and "D" <> 0
ORDER BY "B" desc

如果您希望避免查询中的表达式冗余，这对于更复杂的表达式会派上用场。