SQL select involving multiple tables

Question

In my SQL database, I have a CLIENTS table with an ID column. I have an APPOINTMENTS table containing ID, CLIENTID (which is the ID from the CLIENTS table), and STATUS.

Is there a way to write one SELECT statement that will find the client ID of the client whose STATUS='NO SHOW' the most often? Is there a way to display all clients and ORDER BY how often that an appointment exists where STATUS='NO SHOW'?

Answer 1

Removing the unnecessary join from TomC's answer :

select * from (
    select CLIENTID, count(*) as NoShow 
    from APPOINTMENTS 
    WHERE STATUS='NO SHOW'
    group by CLIENTID
) q order by NoShow desc

Answer 2

select C.ID,A.STATUS from CLIENTS  C
INNER JOIN APPOINEMENTS A ON A.CLIENTID = C.ID
WHERE STATUS='NO SHOW'

Do you mean like this ? Hope this help you

Answer 3

Need to count the number of appointments, and order by that list. Beyond that you can do what you want.

select * from (
    select C.ID, count(*) as NoShow 
    from CLIENTS  C
    INNER JOIN APPOINEMENTS A ON A.CLIENTID = C.ID
    WHERE STATUS='NO SHOW'
    group by c.ID
) q order by NoShow desc

Answer 4

If you want "the most often", then you want to limit the result to one row. The ANSI/ISO SQL construct is fetch first 1 row only :

select CLIENTID
from APPOINTMENTS 
where STATUS = 'NO SHOW'
group by CLIENTID
order by count(*) desc
fetch first 1 row only;

Some databases use limit or select top (1) instead of fetch first .