从SQL中的联接中选择持有最大值的多行

Question

I've read multiple other posts about a similar issue on here, but none of them seem to answer my problem.

I have a table, works_on, which contains all of the hours from a project that employees have worked on:

     id     pno  hours

 123456789   1   32.50
 123456789   2    7.50
 333445555   2   10.00
 333445555   3   10.00 
 333445555  10   10.00
 ...

And another table, project, giving the name of each of the corresponding project numbers (pno) in the works_on table:

   pname           pno  location 

 ProductX           1   Bellaire 
 ProductY           2   Sugarland
 ProductZ           3   Houston
 Computerization   10   Stafford
...

I'm trying to get the name(s) of the project(s) with the highest aggregate hours WITHOUT using a SELECT TOP or LIMIT constraint, as we don't know how many "max" values there are.

I'm able to aggregate the hours as such:

SELECT p.pname, sub.hours FROM company.project p
JOIN
    (SELECT w.pno, SUM(IFNULL(w.hours, 0)) AS hours FROM company.works_on w
     GROUP BY w.pno) AS sub
ON p.pnumber = sub.pno;

Returning the result:

   pname           hours      

 ProductX          52.50
 ProductY          37.50
 ProductZ          50.00
 Computerization   55.00
 Reorganization    25.00
 Newbenefits       55.00

ie, computerization and newbenefits should be ideally returned, as they have the highest aggregate hours, but I can't seem to execute a MAX query that gives me this result.

Is there a workaround to this to select the max result without the use of a TOP or SORT/LIMIT constraint?

Answer 1

This is tough to do in MySQL without the help of analytic functions. Here is one way:

SELECT p.pname, sub.hours
FROM company.project p
INNER JOIN
(
    SELECT pno, SUM(hours) AS hours
    FROM company.works_on
    GROUP BY pno
) sub
    ON p.pnumber = sub.pno
WHERE
    sub.hours = (SELECT SUM(hours) FROM company.works_on
                 GROUP BY pno ORDER BY SUM(hours) DESC LIMIT 1);

If you are using MySQL 8+ or later, then we can take advantage of RANK() :

SELECT pname, hours
FROM
(
    SELECT p.pname, sub.hours,
        RANK() OVER (PARTITION BY p.pname ORDER BY sub.hours DESC) rnk
    FROM company.project p
    INNER JOIN
    (
        SELECT pno, SUM(hours) AS hours
        FROM company.works_on
        GROUP BY pno
    ) sub
) t
WHERE rnk = 1;