SQL 如何在同一个表中具有特定行值的 select 参与者 ID，其中这些行值来自另一个模式中的查找列表？

Question

我想从一个表中 select 参与者在 B 列中同时诊断为癌症和糖尿病

例如：

ParticipantID Diagnosis
1234          Cancer 
1234          Diabetes
4567          Cancer
4567          Lung Disease
8910          Stroke
1256          Lung Disease
3489          Chron's Disease

如何编写一个查询，允许我查询具有两个（或多个）特定诊断的 select 参与者，例如，我只想查看同时患有癌症和糖尿病的参与者？ （在上述示例中，参与者 ID 为 1234）

我尝试做一个简单的 select 语句：

SELECT diagnosisifknown.participantid
diagnosisifknown.diagnsis
diagnosisifknown.date
FROM
diagnosisifknown
where diagnosis ="1" and "10"; 

（查找表中这些诊断的值）

我想要一个像这样的 output ：

1234   Cancer
1234   Diabetes

Answer 1

聚合提供了一种简单的选择：

SELECT ParticipantID
FROM yourTable
GROUP BY ParticipantID
HAVING MIN(Diagnosis) <> MAX(Diagnosis);

你也可以在这里使用存在逻辑：

SELECT t1.ParticipantID
FROM yourTable t1
WHERE EXISTS (SELECT 1 FROM yourTable t2
              WHERE t2.ParticipantID = t1.ParticipantID AND t2.Diagnosis <> t1.Diagnosis);

Answer 2

你可以用 cte 来做，这里是演示。

with cte as
(
  select
    ParticipantID
  from myTable
  where Diagnosis in ('Cancer', 'Diabetes')
  group by
      ParticipantID
  having count(ParticipantID) > 1
)

select
    c.ParticipantID,
    m.Diagnosis
from cte c
join myTable m
on c.ParticipantID = m.ParticipantID

output：

| participantid | diagnosis |
| ------------- | --------- |
| 1234          | Cancer    |
| 1234          | Diabetes  |

Answer 3

如果你想要参与者，你可以使用：

SELECT d.participantid
FROM diagnosisifknown d
WHERE diagnosis IN ('Cancer', 'Diabetes')
GROUP BY d.participantid
HAVING COUNT(*) = 2;

如果您想要原始行，我可能会建议EXISTS ：

select d.*
from diagnosisifknown d
where exists (select 1
              from diagnosisifknown d2
              where d2.participantid = d.participantid and d2.diagnosis = 'Cancer'
             ) and
      exists (select 1
              from diagnosisifknown d2
              where d2.participantid = d.participantid and d2.diagnosis = 'Diabetes'
             ) ;