SQL Server matching all rows from Table1 with all rows from Table2 - sql

Question

someone please help me with this query,
i have 2 tables
Employee
EmployeeID LanguageID
1 1
1 2
1 3
2 1
2 3
3 1
3 2
4 1
4 2
4 3
Task
TaskID LanguageID LangaugeRequired
1 1 1
1 2 0
2 1 1
2 2 1
2 3 1
3 2 0
3 3 1
LangaugeID is connected to table langauge (this table is for explaination only)
LangaugeID LanguageName
1 English
2 French
3 Italian
is there a possilbe way to make a query which gets employees where they can speak all the languages required for each task?
for example:
Task ID 1 requires only LanguageID = 1, so the result should be EmployeeID 1,2,3,4
Task ID 2 requires all 3 languages, so the result should be EmployeeID 1,4
Task ID 3 requires only LanguageID = 3, so the result should be EmployeeID 1,2,4

Answer 1

here is another variant to do this:
select t1.taskid, t2.employeeid from
(
select a.taskid, count(distinct a.languageid) as lang_cnt
from
task as a
where a.LangaugeRequired=1
group by a.taskid
) as t1
left outer join
(
select a.taskid, b.employeeid, count(distinct b.languageid) as lang_cnt
from
task as a
inner join
employee as b
on (a.LangaugeRequired=1 and a.languageid=b.languageid)
group by a.taskid, b.employeeid
) as t2
on (t1.taskid=t2.taskid and t1.lang_cnt=t2.lang_cnt)
###
here you can insert where statement, like:
where t1.taskid=1 and t2.employeeid=1
if such query returns row - this employee can work with this task, if no rows - no
###
order by t1.taskid, t2.employeeid
as you see, this query creates two temporary tables and then joins them.
first table (t1) calculates how many languages are required for each task
second table (t2) finds all employees who has at least 1 language required for task, groups by task/employee to find how many languages can be taken by this employee
the main query performs LEFT JOIN, as there can be situations when no employees can perform task
here is the output:
task employee
1 1
1 2
1 3
1 4
2 1
2 4
3 1
3 2
3 4
update: simpler, but less correct variant, because it will not return tasks without possible employees
select a.taskid, b.employeeid, count(distinct b.languageid) as lang_cnt
from
task as a
inner join
employee as b
on (a.LangaugeRequired=1 and a.languageid=b.languageid)
group by a.taskid, b.employeeid
having count(distinct b.languageid) = (select count(distinct c.languageid) from task as c where c.LangaugeRequired=1 and c.taskid=a.taskid)

Answer 2

Another version using NOT EXISTS
Retrieve all task-employee combinations where a missing language does not exist
SELECT t1.EmployeeId, t2.TaskId
FROM (
SELECT DISTINCT EmployeeID
FROM Employee
) t1 , (
SELECT DISTINCT TaskID
FROM Task
) t2
WHERE NOT EXISTS (
SELECT 1 FROM Task t
LEFT JOIN Employee e
ON e.EmployeeID = t1.EmployeeID
AND e.LanguageID = t.LanguageID
WHERE t.TaskID = t2.TaskID
AND LanguageRequired = 1
AND e.EmployeeID IS NULL
)
http://www.sqlfiddle.com/#!6/e3c78/1

Answer 3

You could use a Join logic to get the result, something like:
SELECT a.EmployeeID FROM Employee a, Task b WHERE b.LanguageRequired == a.LanguageID;