我有公交车、街道、去程、回程的表
在餐桌街我有
示例：

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，公交车101从街道1，2和5按这个顺序行驶，公交车从街道3，2和1按这个顺序返回。
我想知道哪些公交车经过街道“x”和街道“y”（先经过x，后经过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

因此，我用于发现公交车的sql是...（例如，1号和5号街道的过路情况）

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。谁能帮我说一下这个搜索的“最好”查询是什么？