Question

我有一个更复杂的情况，但我想我可以让这个例子更容易回答。假设我正在尝试查询那些在德克萨斯州并且只驾驶蓝色卡车的人。我正在使用Oracle数据库。

我有2个表（State table）和（Truck table）他们加入了人员ID＃

Auto表非常庞大！这是最大的问题，因为我不希望查询挂起或运行很长时间。

州表

 People     State
 ------     -----
 111        TX  
 222        TX  
 333        CA

卡车表

 People     Date     Color
 ------     ----     ------
 111        1995     Orange
 111        1996     Blue
 111        1997     White
 111        2017     Blue
 111        2017     Gold
 222        2006     Blue
 333        2007     Blue

我想要归还的只有222，因为他们没有任何其他颜色的卡车。

 222       2006     Blue

如果有更好的方法来创建看起来像桌子的东西，我也想知道。

Answer 1

如何使用。

select state.people, state,  color, count(color)
from state, truck
where state.people = truck.people
and state.state = 'TX'
and color = 'Blue'
group by state.people, state, color
having count(color) = 1

这里是小提琴

http://sqlfiddle.com/#!4/89bdc/14

Answer 2

假设所有列都有NOT NULL约束：

select State.People
from State
join Truck on Truck.People = State.People
where State.State = 'TX'
and Truck.Color = 'Blue'
and not exists (
  select null
  from Truck t2
  where t2.People = Truck.People
  and t2.Color != 'Blue'
);

Truck (People, Color)上的覆盖索引可能足以满足性能。

另一种考虑因素：

select State.People
from State
join Truck on Truck.People = State.People
where State.State = 'TX'
and Truck.Color = 'Blue'
minus
select Truck.People
from Truck
where Truck.People = Truck.People
and Truck.Color != 'Blue';

为了进行更多性能调优，我将首先查看执行计划和近似数据量。

Answer 3

ADDED （事后）：下面的解决方案只返回people标识符（本例中为222），而不是{{1}的行（或行！）那些人的表。如果需要完整行而不是聚合查询，则可以使用分析函数。易于修改，但首先回答我在原始问题的评论中提出的问题。结束编辑。

truck

Answer 4

如下所示尝试子查询：

select * from (select state.People, truck.Date, truck.Color, 
count(truck.People) over (partition by state.People) Cnt from state
left join truck on state.People = truck.People and truck.Color = 'Blue'
where state.State = 'TX')
where cnt = 1

Answer 5

您可以首先使用JOIN查询表：

SELECT People, Date, Color
  FROM Truck 
    INNER JOIN State ON Truck.People = State.People
  WHERE State.State = 'TX'
    AND Truck.Color = 'Blue'

如果查询性能不足，那么您可能希望在任一表上添加适当的索引。例如，您可能希望在Color属性上为Truck表添加索引。

如何通过代码消除搜索中的人员

5 个答案: