2

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

表 route_going，我有示例：

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

...

表 route_return，我有例子：

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

...

好的，在这个例子中，公共汽车 101 按此顺序从街道 1,2 和 5 行驶。巴士从 3,2 和 1 号街道返回，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？

4

2 回答 2

2 
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;
于 2013-08-01T08:37:18.647 回答
0 
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;

尝试这个。谢谢

于 2013-08-01T06:54:55.523 回答