find the manager of sales employee - mysql

Question

my table is 'DESIGNATION'
ID || DEPT_ID || E_NAME || DESIGNATION
1 || 12 || A || EMPLOYEE
2 || 12 || B || MANAGER
3 || 12 || C || EMPLOYEE
4 || 14 || D || MANGER
5 || 14 || E || EMPLOYEE
6 || 14 || F || EMPLOYEE
I want the manager name through their DEPT_NAME....
Mean result will look like
ID || DEPT_ID || E_NAME || DESIGNATION || MANAGER
1 || 12 || A || EMPLOYEE || B
2 || 12 || B || MANAGER || B
3 || 12 || C || EMPLOYEE || B
4 || 14 || D || MANGER || D
5 || 14 || E || EMPLOYEE || D
6 || 14 || F || EMPLOYEE || D
My query is
SELECT `ID`,`DEPT_ID`,`ENAME`,`DESIGNATION`,
(select `ENAME` from `DESIGNATION` where
(select `E_NAME` from `DESIGNATION` where
(SELECT `DEPT_ID` FROM `DESIGNATION` WHERE `DESIGNATION` = 'EMPLOYEE')
=
(SELECT `DEPT_ID` FROM `DESIGNATION` WHERE `DESIGNATION` = 'MANAGER') and `DESIGNATION`='MANAGER')
AS MANAGER
from `DESIGNATION`
but its not working...

Answer 1

You just need a JOIN operation. This is a basic concept when working with database. You should take some time reading about it.
Something like that?
SELECT A.*, B.E_NAME
FROM DESIGNATION AS A, DESIGNATION AS B
WHERE B.DESIGNATION = "MANAGER"
AND A.DEPT_ID = B.DEPT_ID
Or using explicit JOIN syntax:
SELECT A.*, B.E_NAME
FROM DESIGNATION AS A JOIN DESIGNATION AS B USING (DEPT_ID)
WHERE B.DESIGNATION = "MANAGER"
EDIT:
If you could have multiple managers, you could use the GROUP_CONCAT aggregate functions with explicit group by E_NAME (assuming this is an unique key):
SELECT A.*, GROUP_CONCAT(B.E_NAME)
FROM DESIGNATION AS A, DESIGNATION AS B
WHERE B.DESIGNATION = "MANAGER"
AND A.DEPT_ID = B.DEPT_ID
GROUP BY(A.E_NAME)

Answer 2

Old-school join syntax - sorry - but your sub queries don't make much sense.
Select d.ID,
d.DEPT_ID,
d.E_NAME,
d.DESIGNATION
m.MANAGER
from designation d,
designation m
where d.dept_id = m.dept_id
and m.designation = 'MANAGER'

Answer 3

This looks like a better job for whatever server-side script you're running. For example, in PHP, you can do this:
$managers = []; // array() before version 5.4
$employees = []; // see above comment
$query = "SELECT * FROM `DESIGNATION`";
$result = mysql_query($query); // adjust according to extension of choice
while($row = mysql_fetch_assoc($result)) { // same as previous comment
$employees[] = $row;
if( $row['DESIGNATION'] == "MANAGER") $managers[$row['DEPT_ID']] = $row['E_NAME'];
}
foreach($employees as $i=>$e) {
$employees[$i]['MANAGER'] = $managers[$e['DEPT_ID']] ?: "Nobody";
}

Answer 4

A simple JOIN will help, I prefer an explicit JOIN instead of the implicit comma notation:
SELECT `ID`,`DEPT_ID`,`E_NAME`,`DESIGNATION`, m.e_name, AS `MANAGER`
FROM `DESIGNATION` e
INNER JOIN `DESIGNATION` m
ON e.dept_id = m.dept_id
WHERE m.designation = 'MANAGER'

Answer 5

Can't we simply do this, in more simpler way ?
SELECT ID,DEPT_ID,ENAME,DESIGNATION,
case
when DEPT_ID=12 then 'B'
when DEPT_ID=14 then 'D'
end "MANAGER"
from DESIGNATION

Answer 6

Try the following
SELECT t.id
,t.dept_id
,t.e_name
,t.designation
,ta.e_name As Manager FROM Table1 t JOIN (SELECT e_name,dept_id
FROM Table1 WHERE designation = 'MANAGER'
GROUP BY dept_id,e_name) ta ON ta.dept_id = t.dept_id
SQLFiddle Demo