如何在 Oracle 的表中查找特定值？

Posted 2023-03-27

【中文标题】如何在 Oracle 的表中查找特定值？

【英文标题】：How do I find specific values in a table in Oracle?

【发布时间】：2018-10-05 06:03:14

【问题描述】：

假设我的表是 TEST_123，它有以下记录：

id |  cid | result
------------------
1  |  C-1 |   TAM
2  |  C-1 |   TAM
3  |  C-2 |   RAM
4  |  C-2 |   TAM
5  |  C-3 |   SAM
6  |  C-3 |   SAM

现在我想要这样的 cid，它只有一种类型的结果，所以答案应该是 C-1 AND C-3 但不是 C-2，因为它有两种不同类型的结果。需要 Oracle 查询吗？

【问题讨论】：

欢迎来到 Stack Overflow！提问时请更具体一点：到目前为止，您对代码示例进行了哪些尝试？ (I downvoted because there is no code) / 您有什么期望？ / 您会遇到什么错误？ 请查看“How to ask”以获得帮助

【参考方案1】：

你只需要了解GROUP BY和HAVING子句。

答案很简单

select cid
from TEST_123
group by cid
having count(distinct result) = 1

注意 group by 从CID 中选择不同的键； having 过滤条件适用于组中的所有记录，在您的情况下为count(distinct result) = 1

【讨论】：

谢谢 Marmite .. 虽然它很好地回答了我的查询，但我被 Cid 的 count(result) = 1 卡在了 TEST_123 组中的 select cid 中，并没有区分...但是有一个疑问，当我为 result = TAM 放置 where 子句时，使用这个查询，它给了我 C1 和 C2，但我真正想要的只是 C1，你能理解我吗？

如果您只过滤（使用WHERE）TAM - 那么所有剩余的组每个定义只有一个键（TAM），所以C2in 结果是可以的。如果您正在寻找只有一种结果类型的键，即TAM，则必须制定HAVING count(distinct result) = 1 AND max(result) = 'TAM'（您也可以使用MIN，因为结果只有一个值.

【参考方案2】：

使用存在，它有点棘手，因为每个组的结果应该相同

 select t1.* from TEST_123 t1 where exists(
             select 1 from TEST_123 t2 where t2.cid=t1.cid
                                      and t2.result=t1.result
                                      group by t2.cid,t2.result
                                      having count(*)=
                                       (select count(*) from TEST_123 t3
                                       where t3.cid=t2.cid)
                                      )

示例

with TEST_123 as
(
select 1 as id , 'c-1' as cid , 'tam' as result from dual
union all
select 2 as id , 'c-1' as cid , 'tam' as result from dual
union all
select 3 as id , 'c-2' as cid , 'tam' as result from dual
union all
select 4 as id , 'c-2' as cid , 'ram' as result from dual

)

select distinct t1.cid from TEST_123 t1 where exists(
                 select 1 from TEST_123 t2 where t2.cid=t1.cid
                                          and t2.result=t1.result
                                          group by t2.cid,t2.result
                                          having count(*)=
                                           (select count(*) from TEST_123 t3
                                           where t3.cid=t2.cid)
                                          )

demo

【参考方案3】：

根据@zaynul 的回答，这里有另一个变体：

with TEST_123 as
(
select 1 as id , 'c-1' as cid , 'tam' as result from dual
union all
select 2 as id , 'c-1' as cid , 'tam' as result from dual
union all
select 3 as id , 'c-2' as cid , 'tam' as result from dual
union all
select 4 as id , 'c-2' as cid , 'ram' as result from dual
)
select * from test_123 where cid in (
    select cid from test_123 group by cid having count(distinct result) = 1);

【参考方案4】：

select t.cid from 
(select cid, count(*) as count from table_1 group by cid, result) t 
group by t.cid 
having count(*)=1;

应该适合你

【参考方案5】：

我会使用NOT EXISTS：

SELECT t.*
FROM table t
WHERE NOT EXISTS (SELECT 1 FROM table t1 WHERE t1.cid = t.cid AND t1.result <> t.result);