改善SQL查询以避免合并的另一种方法？

Question

用户可以通过文本框中的邮政编码（例如：L14，L15，L16）或位置进行搜索。

如果用户键入“ Liverpool”，它将找到位于“ Liverpool”的所有商店。 如果用户输入邮政编码（例如：L15），它将搜索所有在L15邮政编码区域中交货的商店。

请参阅下表：

mysql> select * from shops;
+----+----------+-----------+----------+
| id | name     | location  | postcode |
+----+----------+-----------+----------+
|  1 | Shop One | Liverpool | L10      |
|  2 | Shop Two | Liverpool | L16      |
+----+----------+-----------+----------+

--

mysql> select * from shops_delivery_area;
+------------------+---------+----------+---------------+
| delivery_area_id | shop_id | postcode | delivery_cost |
+------------------+---------+----------+---------------+
|                1 |       1 | L10      |          1.50 |
|                2 |       1 | L11      |          0.00 |
|                3 |       1 | L12      |          1.00 |
|                4 |       1 | L13      |          1.00 |
|                5 |       2 | L10      |          2.50 |
|                6 |       2 | L16      |          0.00 |
|                7 |       2 | L28      |          0.00 |
+------------------+---------+----------+---------------+

SQL查询：

SELECT U.* FROM 
   ((SELECT DISTINCT shops.*, DA.delivery_cost, DA.postcode AS AreaPostcode FROM shops
             JOIN shops_delivery_area as DA on (DA.shop_id = shops.id)
   WHERE DA.postcode = "Liverpool")
  UNION
   (SELECT DISTINCT shops.*, DA.delivery_cost, DA.postcode AS AreaPostcode FROM shops
             JOIN shops_delivery_area as DA on  
                              (DA.shop_id = shops.id AND
                              DA.postcode = shops.postcode)
   WHERE shops.location = "Liverpool")) as U

--

结果-按位置（利物浦）：

+----+----------+-----------+----------+---------------+--------------+
| id | name     | location  | postcode | delivery_cost | AreaPostcode |
+----+----------+-----------+----------+---------------+--------------+
|  1 | Shop One | Liverpool | L10      |          1.50 | L10          |
|  2 | Shop Two | Liverpool | L16      |          0.00 | L16          |
+----+----------+-----------+----------+---------------+--------------+

结果-按邮政编码（L12）：

+----+----------+-----------+----------+---------------+--------------+
| id | name     | location  | postcode | delivery_cost | AreaPostcode |
+----+----------+-----------+----------+---------------+--------------+
|  1 | Shop One | Liverpool | L10      |          1.00 | L12          |
+----+----------+-----------+----------+---------------+--------------+

它似乎工作正常...还有其他方法可以缩短SQL查询的时间，以避免union或其他问题吗？

Answer 1

由于所有表和选定的列都相同，因此您只需执行以下操作：

  SELECT DISTINCT shops.*, DA.delivery_cost, DA.postcode AS AreaPostcode FROM shops
             JOIN shops_delivery_area as DA on DA.shop_id = shops.id
   WHERE (DA.postcode = "Liverpool")
      OR (DA.postcode = shops.postcode AND shops.location = "Liverpool")

就像您在迭戈的答案中所说的那样，情况有些不同！ 因此，您可以在WHERE clause补偿该差异。

Answer 2

无论选择什么，请注意短代码并不总是最佳代码。 在许多情况下，如果您的逻辑有足够的分歧，则合并结果确实是最佳的（有时甚至是最干净的编程方式）选项。

就是说，WHERE子句中的以下OR似乎涵盖了您的两种情况...

SELECT DISTINCT
  shops.*,
  DA.delivery_cost,
  DA.postcode AS AreaPostcode
FROM
  shops
INNER JOIN
  shops_delivery_area as DA
    ON (DA.shop_id = shops.id)
WHERE
  (DA.postcode = "Liverpool")
OR
  (DA.postcode = shops.postcode AND shops.location = "Liverpool")

Answer 3

我想念什么？ 你为什么不能

 WHERE DA.postcode = "Liverpool" or shops.location = "Liverpool"

Answer 4

请尝试以下操作：

SELECT DISTINCT shops.*, 
       DA.delivery_cost, 
       DA.postcode 
FROM shops 
       JOIN shops_delivery_area as DA on DA.shop_id = shops.id
WHERE DA.postcode = "Liverpool" 
      OR (location = "Liverpool" and DA.postcode = shops.postcode)