JOIN two SELECT statement results

Question

SELECT t1.ks, t1.[# Tasks], COALESCE(t2.[# Late], 0) AS [# Late]
FROM 
    (SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks) t1
LEFT JOIN
    (SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks) t2
ON (t1.ks = t2.ks);

Try something like this:

SELECT 
* 
FROM
(SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks) t1 
INNER JOIN
(SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks) t2
ON t1.ks = t2.ks

Use UNION:

SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks
UNION
SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks

Or UNION ALL if you want duplicates:

SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks
UNION ALL
SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks

If Age and Palt are columns in the same Table, you can count(*) all tasks and sum only late ones like this:

select ks,
       count(*) tasks,
       sum(case when Age > Palt then 1 end) late
  from Table
 group by ks

you can use the UNION ALL keyword for this.

Here is the MSDN doc to do it in T-SQL http://msdn.microsoft.com/en-us/library/ms180026.aspx

UNION ALL - combines the result set

UNION- Does something like a Set Union and doesnt output duplicate values

For the difference with an example: http://sql-plsql.blogspot.in/2010/05/difference-between-union-union-all.html