在没有 DISTINCT 的 Postgres 中使用 LEFT JOIN 时防止重复行

Question

我有 4 张桌子：

• 项目
• 购买
• 购买物品
• 购买折扣

在这些表中，Purchase Discount 有两个条目，所有其他的只有一个条目。但是当我查询它们时，由于LEFT JOIN，我得到了重复的条目。

此查询将在大型数据库中运行，我听说使用DISTINCT 会降低性能。有没有其他方法可以在不使用DISTINCT 的情况下删除重复项？

这里是SQL Fiddle。

结果显示：

[{"item_id":1,"purchase_items_ids":[1234,1234],"total_sold":2}]

但结果应该是：

[{"item_id":1,"purchase_items_ids":[1234],"total_sold":1}]

Answer 1

核心问题是您的LEFT JOIN 增加了行数。见：

• Two SQL LEFT JOINS produce incorrect result

在加入之前将折扣汇总到单行。或者使用（不相关的）子查询表达式：

SELECT json_agg(items)
FROM  (
   SELECT pi.item_id
        , array_agg(pi.id) AS purchase_items_ids
        , sum(pi.sold) AS total_sold
        ,(SELECT COALESCE(sum(pd.discount_amount), 0)
          FROM   purchase_discounts pd
          WHERE  pd.purchase_id = 200) AS discount_amount
   FROM   purchase_items pi
   WHERE  pi.purchase_id = 200
   GROUP  BY 1
   ) AS items;

结果：

[{"item_id":1,"purchase_items_ids":[1234],"total_sold":1,"discount_amount":12}]

db小提琴here

我添加了一些额外的改进：

• 假设 FK constraints 强制执行参照完整性，我们根本不需要涉及表 purchase 和 items。

• 删除了一个不做任何事情的子查询级别。

• 使用json_agg() 代替array_to_json(array_agg())。

• 将COALESCE() 添加到输出0 或NULL 以无折扣。

由于折扣适用于模型中的购买，而不适用于单个商品，因此为每个商品输出discount_amount 没有意义。考虑使用此查询来返回一个项目数组和一个单独的 discount_amount：

SELECT json_build_object(
         'items'
       , json_agg(items)
       , 'discount_amount'
       , (SELECT COALESCE(sum(pd.discount_amount), 0)
          FROM   purchase_discounts pd
          WHERE  pd.purchase_id = 200)
       )
FROM  (
   SELECT pi.item_id
        , array_agg(pi.id) AS purchase_items_ids
        , sum(pi.sold) AS total_sold
   FROM   purchase_items pi
   WHERE  pi.purchase_id = 200
   GROUP  BY 1
   ) AS items;

结果：

{"items" : [{"item_id":1,"purchase_items_ids":[1234],"total_sold":1}], "discount_amount" : 12}

db小提琴here

使用json_build_object() 组装JSON 对象。

您购买单件商品的示例并不太具有启发性。我为我的小提琴添加了一个包含多个项目且没有折扣的购买。

Answer 2

首先我建议从没有理由存在的查询中删除INNER JOIN items i ON i.id = t.item_id。

然后左加入 Purchase_Discounts 表使用子查询来获取 Discount_amount（如 Lukasz Szozda 的回答中所述）

如果任何产品没有折扣，则Discount_amount 列将显示NULL。如果你想避免它，那么你可以使用COALESCE()，如下所示：

COALESCE(SUM((select sum(discount_amount) from purchase_discounts 
                 where purchase_discounts.purchase_id = purchase.id)),0) as discount_amount

Db-Fiddle：

  SELECT array_to_json(array_agg(p_values)) FROM 
     ( 
       SELECT t.item_id, t.purchase_items_ids, t.total_sold, t.discount_amount FROM 
         ( 
           SELECT purchase_items.item_id AS item_id,
                  ARRAY_AGG(purchase_items.id) AS purchase_items_ids,
                  SUM(purchase_items.sold) as total_sold,
                  SUM((select sum(discount_amount) from purchase_discounts 
                      where purchase_discounts.purchase_id = purchase.id)) as discount_amount
                   FROM items
                   INNER JOIN purchase_items ON purchase_items.item_id = items.id
                   INNER JOIN purchase ON purchase.id = purchase_items.purchase_id              
                  WHERE 
                   purchase.id = 200
                  GROUP by 
                   purchase_items.item_id
         ) as t 
       
     ) AS p_values;

输出：

array_to_json
[{"item_id":1,"purchase_items_ids":[1234],"total_sold":1,"discount_amount":12}]

db小提琴here

Answer 3

我认为左连接不会导致，因为 Inner Join 查询结果与左连接相同，在使用 purchase_id=200 查询的折扣中，您可以使用 row_number 和 partion_by 相同的 2 个结果：

ROW_NUMBER() OVER(PARTITION BY purchase_items.id order by purchase_items.id) rn

然后选择 rn=1。 您更改 sum 函数的查询，我认为您可以使用 from partion_by。

Answer 4

使用相关子查询而不是 LEFT JOIN：

SELECT array_to_json(array_agg(p_values)) FROM 
( 
  SELECT t.item_id, t.purchase_items_ids, t.total_sold, t.discount_amount FROM 
    ( 
      SELECT purchase_items.item_id AS item_id,
             ARRAY_AGG(purchase_items.id) AS purchase_items_ids,
             SUM(purchase_items.sold) as total_sold,
             SUM((SELECT SUM(pd.discount_amount) FROM purchase_discounts pd
                  WHERE pd.purchase_id = purchase.id)) as discount_amount
       FROM items
       INNER JOIN purchase_items ON purchase_items.item_id = items.id
       INNER JOIN purchase ON purchase.id = purchase_items.purchase_id
       WHERE purchase.id = 200
       GROUP by purchase_items.item_id
    ) as t 
  INNER JOIN items i ON i.id = t.item_id 
) AS p_values;

db<>fiddle demo

输出：

[{"item_id":1,"purchase_items_ids":[1234],"total_sold":1,"discount_amount":12}]

Answer 5

LEFT JOIN 不会导致您的重复，我理解您为什么需要它，因为可能没有任何折扣，但是对于提供的数据更改为内部连接会产生相同的结果。因为您使用ARRAY_AGG(purchase_items.id)，您收到重复条目。此外，在提供数据的情况下，表item 和purchase 是不必要的。可以使用sum和distinct on的window版本，减少purchase_id的重复，消除上面提到的表格。最后中间的select ... ) t可以完全去掉。结果：（见演示）

select array_to_json(array_agg(p_values)) 
  from (select distinct on (pi.item_id, pi.id)
                pi.item_id
              , pi.id purchase_items_ids
              , sum(pi.sold) over (partition by pi.item_id) total_sold         
              , sum(pd.discount_amount) over(partition by  pi.item_id)  discount_amount
           from purchase_items pi  
           left join purchase_discounts pd 
             on pd.purchase_id = pi.purchase_id 
          order by pi.item_id, pi.id           
       ) as p_values; 

Answer 6

如果您只能在purchase_discounts 表中拥有多个值，那么将多个purchase_discounts 行聚合到一个之前连接的子查询可以解决问题：

SELECT array_to_json(array_agg(p_values)) FROM 
( 
  SELECT t.item_id, t.purchase_items_ids, t.total_sold, t.discount_amount FROM 
    ( 
      SELECT purchase_items.item_id AS item_id,
             ARRAY_AGG(purchase_items.id) AS purchase_items_ids,
             SUM(purchase_items.sold) as total_sold,
             X.discount_amount
             FROM items
              INNER JOIN purchase_items ON purchase_items.item_id = items.id
              INNER JOIN purchase ON purchase.id = purchase_items.purchase_id
              LEFT JOIN (SELECT purchase_id, sum(purchase_discounts.discount_amount) AS discount_amount FROM purchase_discounts GROUP BY purchase_id) X ON X.purchase_id = purchase.id
             WHERE 
              purchase.id = 200
             GROUP by 
              purchase_items.item_id,
              X.discount_amount
    ) as t 
  INNER JOIN items i ON i.id = t.item_id 
) AS p_values;