PostgreSQL 求和而不是串联

Question

我有以下查询：

SELECT DISTINCT
  branch.id,
  branch.name,
  SUM(CAST(COALESCE(NULLIF(REPLACE(dia_anterior.col_9, ',', '.'), ''), '0.0') AS double precision)) AS total_venta,
  SUM(CAST(COALESCE(NULLIF(REPLACE(dia_anterior.col_4, ',', '.'), ''), '0.0') AS double precision)) AS total_personas,
  SUM(CAST(COALESCE(NULLIF(REPLACE(ocupacion_dia_anterior.col_3, ',', '.'), ''), '0') AS double precision)) AS total_ocupacion
FROM branches AS branch
  INNER JOIN queries AS q_dia_anterior
    ON q_dia_anterior.query_structure_id = (SELECT id FROM query_structures WHERE query_structures.slug = 'dia-anterior')
  INNER JOIN queries AS q_ocupacion_dia_anterior
    ON q_ocupacion_dia_anterior.query_structure_id = (SELECT id FROM query_structures WHERE query_structures.slug = 'ocupacion-dia-anterior')
  INNER JOIN queries AS q_ventas_x_articulo_dia_anterior
    ON q_ventas_x_articulo_dia_anterior.query_structure_id = (SELECT id FROM query_structures WHERE query_structures.slug = 'ventas-x-articulo-dia-anterior')
  INNER JOIN branch_data AS dia_anterior
    ON dia_anterior.query_id = q_dia_anterior.id
      AND dia_anterior.branch_id = branch.id
  INNER JOIN branch_data AS ocupacion_dia_anterior
    ON ocupacion_dia_anterior.query_id = q_ocupacion_dia_anterior.id
      AND ocupacion_dia_anterior.branch_id = branch.id
  INNER JOIN branch_data AS ventas_x_articulo_dia_anterior
    ON ventas_x_articulo_dia_anterior.query_id = q_ventas_x_articulo_dia_anterior.id
      AND ventas_x_articulo_dia_anterior.branch_id = branch.id
WHERE
  branch.account_id = 1
GROUP BY
  branch.id,
  branch.name
ORDER BY
  2

当我对结果进行分组以对列求和时，看起来不同行的值正在串联，从而产生巨大的值。

我是不是漏掉了什么？
为什么我得到多次求和的值？

Answer 1

虽然我认为这不能正确回答问题，但我通过以下查询得到了想要的结果：

WITH dia_actual AS (
SELECT
  branch_data.branch_id,
  CAST(COALESCE(NULLIF(REPLACE(branch_data.col_9, ',', '.'), ''), '0') AS double precision) AS total_venta,
  CAST(COALESCE(NULLIF(REPLACE(branch_data.col_4, ',', '.'), ''), '0') AS double precision) AS total_personas
FROM branch_data
  INNER JOIN queries AS query
    ON query.id = branch_data.query_id
  INNER JOIN query_structures AS query_structure
    ON query_structure.id = query.query_structure_id
WHERE
  query_structure.slug = 'dia-actual'
), ocupacion AS (
SELECT
  branch_data.branch_id,
  CAST(COALESCE(NULLIF(REPLACE(branch_data.col_3, ',', '.'), ''), '0') AS double precision) AS total_ocupacion
FROM branch_data
  INNER JOIN queries AS query
    ON query.id = branch_data.query_id
  INNER JOIN query_structures AS query_structure
    ON query_structure.id = query.query_structure_id
WHERE
  query_structure.slug = 'ocupacion-dia-anterior'
)
SELECT
  branch.name,
  SUM(dia_actual.total_venta) AS total_venta,
  SUM(dia_actual.total_personas) AS total_personas,
  SUM(ocupacion.total_ocupacion) AS total_ocupacion
FROM branches AS branch
  INNER JOIN dia_actual
    ON dia_actual.branch_id = branch.id
  INNER JOIN ocupacion
    ON ocupacion.branch_id = branch.id
WHERE
  branch.account_id = 1
GROUP BY
  branch.name
ORDER BY
  branch.name

使用 CTE 可以正确求和值。

Answer 2

填写一些有根据的猜测，这可能是您想要的查询：

SELECT b.name
     , COALESCE(dia_actual.total_venta   , 0) AS total_venta
     , COALESCE(dia_actual.total_personas, 0) AS total_personas
     , COALESCE(ocupacion.total_ocupacion, 0) AS total_ocupacion
FROM   branches b
LEFT   JOIN LATERAL (
   SELECT sum(replace(NULLIF(bd.col_9, ''), ',', '.')::float) AS total_venta
        , sum(replace(NULLIF(bd.col_4, ''), ',', '.')::float) AS total_personas
   FROM   branch_data      bd
   JOIN   queries          q  ON q.id = bd.query_id
   JOIN   query_structures qs ON qs.id = q.query_structure_id
   WHERE  bd.branch_id = b.id  -- lateral reference
   AND    qs.slug = 'dia-actual'
   ) dia_actual ON true
LEFT   JOIN LATERAL (
   SELECT sum(replace(NULLIF(bd.col_3, ''), ',', '.')::float) AS total_ocupacion
   FROM   branch_data      bd
   JOIN   queries          q  ON q.id = bd.query_id
   JOIN   query_structures qs ON qs.id = q.query_structure_id
   WHERE  bd.branch_id = b.id  -- lateral reference
   AND    qs.slug = 'ocupacion-dia-anterior'
   ) ocupacion ON true
WHERE  b.account_id = 1
ORDER  BY  b.name;

备注

先求和，再连接，避免因代理交叉连接导致的行相乘。
- Two SQL LEFT JOINS produce incorrect result
使用 LEFT JOIN 避免丢失行，因为在右侧的一个查询中找不到值。
但是，既然你 select 一个 branch_id 我切换到 LEFT JOIN LATERAL ... ON true，假设你只使用所有行的一小部分。
不需要更昂贵的 CTE。
还简化了从 text 到 double precision 的转换。

当然，最好先将这些数字存储为 numeric data type，而不是 text。

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备注