Question

我在postgres数据库中有多个表，它们包含完全独特的信息。当在查询中正确连接在一起时，信息将产生我正在寻找的所有可能的组合。我正在寻找的信息是完整的SKU。

要生成完整的SKU，此查询会生成所需的结果： 功能查询

SELECT
    materials.code,
    "part_base_parts".code as part_base_parts_id,
    shanks.code AS shank_id,
    measurements.description
FROM
    "part_base_parts"
LEFT JOIN "part_types" ON "part_base_parts"."part_type_id" = "part_types"."id"
RIGHT JOIN "parts_to_shanks" ON "part_base_parts"."id" = "parts_to_shanks"."part_base_part_id"
RIGHT JOIN "parts_to_measurements" ON "part_base_parts"."id" = "parts_to_measurements"."part_base_part_id"
RIGHT JOIN "parts_to_materials" ON "part_base_parts"."id" = "parts_to_materials"."part_base_part_id"
JOIN materials ON "parts_to_materials"."material_id" = materials."id"
JOIN shanks ON "parts_to_shanks"."shank_id" = shanks."id"
JOIN measurements ON "parts_to_measurements"."measurement_id" = measurements."id"
ORDER BY
    part_base_parts_id ASC,
    materials.code ASC,
    shank_id ASC,
    measurements.description ASC

鉴于此查询，我生成了32,640条记录（未应用索引），查询时间为0.82秒。像这样......

给定输出

code    part_base_parts_id  shank_id    description
AA      5105                A           03.0
.
. 32,638 rows in here.
.
ST      6939                D           9/16

但是，这只会让我走到一半。我需要从查询中取回结果并生成每列的计数总数。所以我需要的结果是：

期望的结果

code: AA - ###0
      ...
      ST - ###0
part_base_parts_id: 5105 - ###0
                    ...
                    6939 - ###0
shank_id: A - ###0
          ...
          D - ###0
description: 03.0 - ###0
             ...
             9/16 - ###0

有没有办法从Postgres产生“期望的结果”？

Answer 1

如果你想要它们的行，那么肯定。

WITH cte AS(    
    SELECT
        materials.code,
        "part_base_parts".code as part_base_parts_id,
        shanks.code AS shank_id,
        measurements.description
    FROM
        "part_base_parts"
    LEFT JOIN "part_types" ON "part_base_parts"."part_type_id" = "part_types"."id"
    RIGHT JOIN "parts_to_shanks" ON "part_base_parts"."id" = "parts_to_shanks"."part_base_part_id"
    RIGHT JOIN "parts_to_measurements" ON "part_base_parts"."id" = "parts_to_measurements"."part_base_part_id"
    RIGHT JOIN "parts_to_materials" ON "part_base_parts"."id" = "parts_to_materials"."part_base_part_id"
    JOIN materials ON "parts_to_materials"."material_id" = materials."id"
    JOIN shanks ON "parts_to_shanks"."shank_id" = shanks."id"
    JOIN measurements ON "parts_to_measurements"."measurement_id" = measurements."id"
    ORDER BY
        part_base_parts_id ASC,
        materials.code ASC,
        shank_id ASC,
        measurements.description ASC    
)
SELECT key, value, count(*)
FROM(
  SELECT 'code' AS key, code AS value
  FROM cte
  UNION ALL
  SELECT 'part_base_parts_id', code
  FROM cte
  UNION ALL
  SELECT 'shank_id', shank_id
  FROM cte
  UNION ALL
  SELECT 'description',  description
  FROM cte
) AS q
GROUP BY key, value
ORDER BY key, value

聚合表中的行数

1 个答案: