Display column values next to each other SQL - sql

Question

I have to select which wards have patients with the same last name. I need to display the name of those patients and the IDs that come with them.
My query below works to give me the proper result:-
SELECT pt.NAME, pt.PATIENTID, pt.PATIENTID
FROM PATIENT pt
WHERE pt.NAME IN
(SELECT pt.NAME
FROM PATIENT pt
GROUP BY pt.NAME
HAVING COUNT(DISTINCT pt.WARDNO) < 2)
AND NAME IN
(SELECT NAME
FROM PATIENT
GROUP BY NAME
HAVING COUNT(*) > 1)
It gives me the following result in the table:
Name | PatientID
Jones | p10
Jones | p29
However, I want to display my results in the table like this and am not sure how.
Name | PatientID | PatientID
Jones | p10 | p29

Answer 1

Assuming you expect no more than two patient ID values, then a simple aggregation can work here:
WITH cte AS (
SELECT NAME
FROM PATIENT
GROUP BY NAME
HAVING COUNT(*) > 1 AND COUNT(DISTINCT WARDNO) < 2
)
SELECT
NAME,
MIN(PATIENTID) AS PatientID1,
MAX(PATIENTID) AS PatientID2
FROM cte
GROUP BY
NAME;

Answer 2

Thank you Tim. That aggregation did work and I did a little tweak to make it a lot cleaner. It works to display just the two PatientIDs which is what I need for now.
My new query is as follows
SELECT pt.NAME, MIN(pt.PATIENTID) AS PAT, MAX(pt.PATIENTID) AS PAT
FROM PATIENT pt
GROUP BY pt.NAME
HAVING COUNT(*) > 1 AND COUNT(DISTINCT pt.WARDNO) < 2
However, if I do have multiple PatientIDs I am curious what I would do.

Answer 3

I have to select which wards have patients with the same last name.
Assuming patients are not duplicated, you can get the patient/ward combinations as:
select p.wardno, p.name, count(*)
from patient p
group by p.wardno, p.name
having count(*) >= 2;
If you only want the wards, then you have one of the rare situations where select distinct and group by are appropriate:
select distinct p.wardno
from patient p
group by p.wardno, p.name
having count(*) >= 2;