对此最好的查询是什么？答案

【问题标题】：What is the best query for this?对此最好的查询是什么？
【发布时间】：2013-08-01 22:49:08
【问题描述】：

我有表 bus, street, route_going, route_return

在我的餐桌街

例子：

id | name
1  | street1 
2  | street2
3  | street4
...
n  | streetn

table_route_going，我有例子：

id_bus | id_street | order
101    | 1         | 1
101    | 2         | 2
101    | 5         | 3

...

table route_return，我有例子：

id_bus | id_street | order
101    | 3         | 1
101    | 2         | 2
101    | 1         | 3

...

好的，在此示例中，公共汽车 101 按此顺序从街道 1、2 和 5 行驶。以及从 3,2 和 1 号街道返回的公共汽车，按此顺序。

我想知道哪些公共汽车经过街道“x”和街道“y”（先是 x，后是 y）

例如：

x = 1, y = 5 -> the bus 101 pass
x = 1, y = 3 -> the bus 101 pass
x = 3, y = 1 -> the bus 101 pass
x = 3, y = 5 -> the bus 101 don't pass

所以，我用于发现公共汽车的 sql 是...（例如通过街道 1 和 5）

select * from bus as b where 
-- The bus passes between the 2 streets at the going route??
exists (select * from route_going as rg1, route_going as rg2,street as r1,street as r2 where rg1.id_bus = rg2.id_bus and rg1.id_street = r1.id and rg2.id_street = r2.id and r1.id = 1 and r2.id = 5 and b.bus_id = rg1.id_bus and rg1.order <= rg2.order)
-- The bus passes between the 2 streets at the return route??
or exists (select * from route_return as rg1, route_return as rg2,street as r1,street as r2 where rg1.id_bus = rg2.id_bus and rg1.id_street = r1.id and rg2.id_street = r2.id and r1.id = 1 and r2.id = 5 and b.bus_id = rg1.id_bus and rg1.order <= rg2.order)
-- The bus passes between the 2 streets at the going route first and return route later??
or exists (select * from route_going as rg1, route_return as rg2,street as r1,street as r2 where rg1.id_bus = rg2.id_bus and rg1.id_street = r1.id and rg2.id_street = r2.id and r1.id = 1 and r2.id = 5 and b.bus_id = rg1.id_bus)

所以，我认为这个查询不好。有人可以帮我说出这个搜索的“最佳”查询吗？

【问题讨论】：

我们可以有一个SQL Fiddle 吗？请？此外，添加有关索引的信息。这可能会有所作为。
我不明白，你的代码做的事情太多了，只是检查公共汽车是否穿过街道
@NoIdeaForName 穿过两条条街道！按顺序！
@jpmc26 sqlfiddle.com/#!2/a0a7d/7
我认为真正使问题复杂化的部分是订购要求。如果问题很简单，“它穿过这两条街道吗？”它会简单得多。如果去和返回不使用相同的数字进行订购也会更简单。（如，如果返回值总是更高。）我最初的想法是将两个路线表合并到一个表中，并添加一列来指示公共汽车是否正在/返回，并在该表中使用数字将完全整理路线中的街道。

标签： mysql sql database select exists

【解决方案1】：

SET @x = 1;
SET @y = 5;

SELECT x.id_bus
  FROM   
     ( SELECT *,0 returning FROM route_going UNION SELECT *, 1 FROM route_return ) x
  JOIN
     ( SELECT *,0 returning FROM route_going UNION SELECT *, 1 FROM route_return ) y
    ON y.id_bus = x.id_bus
   AND (y.returning > x.returning OR (y.returning = x.returning AND y.porder > x.porder))
 WHERE x.id_street = @x
   AND y.id_street = @y;

【讨论】：

如果您想要 DISTINCT 结果，请使用 DISTINCT 运算符。
我想知道UNION ALL 是否会带来任何性能提升。因为这在UNION 中指定了0 或1，所以不会有相同的行，因此检查重复行（如果引擎执行它）是浪费循环。

【解决方案2】：

select 
street.name,
route_going.id_bus,route_going.id_street
route_return.id_bus,route_return.id_street
on street.id=route_going.id_street and 
route_return.id_street=route_going.id_street;

试试这个。谢谢

【讨论】：