Mysql 另一种改进SQL查询以避免联合的方法？

用户可以通过邮政编码（例如：L14、L15、L16）或文本框中的位置进行搜索

如果用户输入“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邮政编码区域内送货的商店

见下表：

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

-

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

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

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

-

结果-按地点（利物浦）：

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

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

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

结果-按邮政编码（L12）：

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

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

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

它似乎工作正常。。。 有没有其他方法可以缩短SQL查询以避免

联合

或其他问题？

我缺少什么？ 你为什么不能呢

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

---

由于所有表和选定列都是相同的，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")

就像你在迭戈的回答中所说的，情况有点不同！因此，您可以在

WHERE子句中补偿该差异，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)

无论您选择什么，请注意短代码并不总是最佳代码。在许多情况下，当您有足够的分歧逻辑时，联合结果确实是最理想的（有时也是最干净的，编程上的）选择

也就是说，以下或WHERE条款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")

这将不能正常工作，否则您将得到许多行相同的商店名称。仔细查看
JOIN
。根据索引的使用方式，使用
或
会降低性能。我通常会像这样将查询拆分为
联合
，以避免出现
或
，并通过在
联合
的每个部分中使用一个索引来获得可观的性能增益@Diego，当它是
shops.location=“Liverpool”时，您缺少一个限制
。只需在位置测试中添加另一个约束条件，以避免类似DA.postcode=shops.postcode的情况。您有什么理由避免使用
联合
？p、 我想你可以删除
DISTINCT
关键字，因为
UNION
是
UNION DISTINCT
的缩写。谢谢！那么，在性能方面，哪一个是更好的选择呢？坚持使用Union或
或
中的WHERE子句，我刚刚做了性能测试。。使用
Union
似乎更快。@user791022 said
我刚刚做了性能测试。。使用Union似乎更快。
我早就告诉过你了。在一个表的中间进行两次索引命中或扫描整个表，什么速度更快？当您对SQL应用“应用程序”思想（减少冗余代码等）时，您会遇到性能问题。@KM.-虽然我在技术细节上同意你的观点，但我也只是在被证明是错误的情况下才学到了这个教训。随着我对SQL知识的增长，我寻找创新的方法来缩短代码。那时我发现了执行计划以及它们与我对SQL的理解之间的关系。我认为OP一直在寻找更好的方法是件好事。我甚至很高兴OP挑战了其他人的说法，解决了与理论和其他人所说的不符的证据。