ORA-00933 when trying to compare counts - sql

Question

This is the question
List the name of employee who work on more projects than employee 'Grace'', Show three columns in result: name of employee, project count of employee, grace's project count.
This is my code
SELECT employee."NAME", T1."# OF PROJECTS",
(SELECT COUNT(pid) FROM workon WHERE empid = 30) AS "Grace's Project"
FROM employee,
(SELECT empid, COUNT(pid) AS "# OF PROJECTS"
FROM workon
GROUP BY empid
ORDER BY empid)AS T1
WHERE T1."# OF PROJECTS" > (SELECT COUNT(pid) FROM workon WHERE empid = 30)
AND t1.empid = employee.EMPID
I keep getting ORA-00933: SQL command not properly ended. what am I missing?

Answer 1

The only error in your query is that Oracle does not accept AS for table aliases. Remove it and your query runs just fine.
There are two things I'd like to mention, though:
You are using an ancient join syntax you shouldn't use anymore. Comma-separated joins were made redundant by the introduction of explicit joins (e.g. INNER JOIN ... ON) in 1992.
Your query is a little over-complicated. Most of all, because you are counting projects thrice, once for all employees, twice for Grace. You can avoid this by using WITH clauses.
Here is the query built-up step by step with WITH clauses:
WITH emp AS
(
SELECT empid, e.name, COUNT(*) AS projects
FROM workon w
JOIN employee e USING(empid)
GROUP BY empid, e.name
ORDER BY empid
)
, grace AS
(
SELECT * FROM emp WHERE name = 'Grace'
)
SELECT
emp.name,
emp.projects as "# OF PROJECTS",
grace.projects as "Grace's Projects"
FROM emp
CROSS JOIN grace
WHERE emp.projects > grace.projects
ORDER BY emp.projects DESC, emp.name;
Demo: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=f40e2c33541c76f0af112be967370784

Answer 2

I would use window functions:
select ew.name, ew.num_projects
from (select e.empid, e."NAME", count(*) as num_projects,
max(case when ew."NAME" = 'Grace' then count(*) else 0 end) over () as grace_num_projects
from employee e join
workon w
on w.empid = e.empid
group by e.empid, e."NAME"
) ew
where num_projects > grace_num_projects;

Answer 3

SELECT DISTINCT
e.empid,
,e."NAME"
FROM employee e
INNER JOIN (SELECT
empid
,count(*) as num_projects
FROM workon
GROUP BY empid) w ON w.empid = e.empid
LEFT JOIN (SELECT
1 AS ID
COUNT(*) as grace_projects
FROM workon
GROUP BY empid
WHERE empid = 30) g ON g.ID = 1
WHERE w.num_projects > g.grace_projects;
So here's what I am doing.
I am counting the projects before they are joined, that should decrease the overhead for the query, as the return on the join is shrank considerably prior to joining.
An index on that WORKON table by EmpID would speed up the query considerably.
Then, I query Graces figures, because I want them to return a value against any person, it should only return one result for a count, and then just join that by an arbitrary value so it returns against all rows
Again, it should utilise the same index.
Because it is calculating Graces figures first, it should only need to do this once, whereas a subquery would need calculate graces figures for each employee which is an unnecessary overhead.
This is then filtered in the where clause.