[Update]
Thanks for all the comments!
Appreciated!
I solved this by the code below, after referring to all of your posts.
I did not put into account of hierarchy but I will for the later works.
Thank you!
SELECT m.first_name
, m.last_name
, RANK() over (partition by (select COUNT(e.employee_id)
from DB1_employee e)
ORDER BY e.employee_id DESC) AS RANK from DB2_manager m LEFT JOIN RITDB_employee e ON m.employee_id=e.manager
I am having difficulty to troubleshoot my code to rank the managers
based on the number of employees they have. The error is "missing right parenthesis". Any helps would be highly appreciated. Thanks!
SELECT m.first_name
, m.last_name
, RANK() over (partition by (select COUNT(e.employee_id)
from DB1_employee e) AS NUM_EMP
ORDER BY NUM_EMP DESC) AS RANK
from DB2_manager m
, DB1_employee e
group by m.first_name, m.last_name
ORDER BY RANK
An example based on Scott's schema (as I don't have your tables nor data), so that it is easier to see what should be returned.
SQL> select mgr, empno, ename, job from emp order by mgr;
MGR EMPNO ENAME JOB
---------- ---------- ---------- ---------
7566 7902 FORD ANALYST --> MGR 7566 (Jones) has 2 employees
7566 7788 SCOTT ANALYST
7698 7900 JAMES CLERK --> 7698 (Blake) has 5 employees
7698 7499 ALLEN SALESMAN
7698 7521 WARD SALESMAN
7698 7844 TURNER SALESMAN
7698 7654 MARTIN SALESMAN
7782 7934 MILLER CLERK --> 7782 (Clark) has 1 employee
7788 7876 ADAMS CLERK --> 7788 (Scott) has 1 employee
7839 7698 BLAKE MANAGER --> 7839 (King) has 3 employees
7839 7566 JONES MANAGER
7839 7782 CLARK MANAGER
7902 7369 SMITH CLERK --> 7902 (Ford) has 1 employee
7839 KING PRESIDENT
14 rows selected.
SQL>
So:
SQL> select m.ename mgrname,
2 count(*) cnt,
3 dense_rank() over (order by count(*) desc) rnk
4 from emp e join emp m on e.mgr = m.empno
5 where e.mgr is not null
6 group by e.mgr, m.ename
7 order by rnk;
MGRNAME CNT RNK
---------- ---------- ----------
BLAKE 5 1
KING 3 2
JONES 2 3
SCOTT 1 4
FORD 1 4
CLARK 1 4
6 rows selected.
SQL>
However:
That's not correct. Why? Because of hierarchy. Not all managers share the same level. Have a look: King is the president; he is the big boss and manages them all; how can he be ranked as #2? The same goes for the rest of them.
SQL> select lpad(' ', 2 * level - 2) || e.empno ||' '|| e.ename val
2 from emp e
3 connect by prior e.empno = e.mgr
4 start with e.mgr is null;
VAL
--------------------------------------------------------------------
7839 KING
7566 JONES --> Jones has 4 employees!!! (Scott, Adams, Ford and Smith)
7788 SCOTT --> Scott has 1 employee (that's Adams)
7876 ADAMS
7902 FORD --> Ford has 1 employee (that's Smith)
7369 SMITH
7698 BLAKE
7499 ALLEN
7521 WARD
7654 MARTIN
7844 TURNER
7900 JAMES
7782 CLARK
7934 MILLER
14 rows selected.
SQL>
Therefore, we need another approach. Begin with a simple query which, basically, returns "root" for each of them:
SQL> select connect_by_root(ename) manager
2 from emp
3 connect by prior empno = mgr;
MANAGER
----------
SCOTT
SCOTT
FORD
FORD
ALLEN
JAMES
<snip>
KING
KING
39 rows selected.
SQL>
It is further used as a source for
SQL> select x.mgrname,
2 count(*) - 1 cnt
3 from (select connect_by_root(e.ename) mgrname
4 from emp e
5 connect by prior e.empno = e.mgr
6 ) x
7 group by x.mgrname;
MGRNAME CNT
---------- ----------
ALLEN 0
JONES 4
FORD 1
MILLER 0
CLARK 1
WARD 0
SMITH 0
SCOTT 1
TURNER 0
MARTIN 0
ADAMS 0
JAMES 0
BLAKE 5
KING 13
14 rows selected.
SQL>
Finally, remove those that don't have any employees and rank them; the result is quite different than the first (the most obvious, but probably wrong) approach:
SQL> select r.mgrname,
2 r.cnt,
3 dense_rank() over (order by cnt desc) rnk
4 from (select x.mgrname,
5 count(*) - 1 cnt
6 from (select connect_by_root(e.ename) mgrname
7 from emp e
8 connect by prior e.empno = e.mgr
9 ) x
10 group by x.mgrname
11 ) r
12 where r.cnt > 0
13 order by rnk;
MGRNAME CNT RNK
---------- ---------- ----------
KING 13 1
BLAKE 5 2
JONES 4 3
SCOTT 1 4
FORD 1 4
CLARK 1 4
6 rows selected.
SQL>
First of all
Your query is doing a cross join, and then grouping by manager name. There can be an issue if two managers have the same name. Better to use PK and FK.
Don't use old join syntax
NUM_EMP. has no reference in your query and still used in the ORDER BY.
I am considering that DB1_employee table must have some column referring to its manager(let's say, MANAGER_FK ). So writing the query accordingly as following:
SELECT
M.FIRST_NAME,
M.LAST_NAME,
RANK() OVER(
ORDER BY
E.EMPLOYEE_CNT DESC
) AS RNK
FROM
DB2_MANAGER M
JOIN (
SELECT
MANAGER_FK,
COUNT(1) AS EMPLOYEE_CNT
FROM
DB1_EMPLOYEE
GROUP BY
MANAGER_FK
) E ON ( M.MANAGER_ID = E.MANAGER_FK )
ORDER BY
RNK;
Cheers!!
try:
SELECT m.first_name
, m.last_name
, RANK() over (partition by (select COUNT(e.employee_id)
from DB1_employee e)
ORDER BY NUM_EMP DESC) AS RANK
from DB2_manager m
, DB1_employee e
group by m.first_name, m.last_name
ORDER BY RANK
;
Related
I want to fetch data based on the last row number.
I want the records highlighted in yellow color. Please guide.
Should be a simple MAX function.
select max(row_number) rn,
account_no
from your_table
group by account_no
order by account_no;
If "row_number" represents the result of analytic function (which isn't clear from what you posted so far), then include ORDER BY clause into the function (I don't know which column you're sorting data on) in descending order so that your "max" actually becomes "min" whose RN = 1 and then it is easy to select it as a final result.
with temp as
(select columnb,
columnc,
row_number() over (partition by accountno order by SOMETHING desc) rn
^^^^^^^^^^^^^^^^^^^^^^^
add this
from some_table
)
select columnb,
columnc
from temp
where rn = 1
As I don't have your tables, here's Scott's EMP:
SQL> select * from emp;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- -------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17.12.80 920 20
7499 ALLEN SALESMAN 7698 20.02.81 1600 300 30
7521 WARD SALESMAN 7698 22.02.81 1250 500 30
7566 JONES MANAGER 7839 02.04.81 2975 20
7654 MARTIN SALESMAN 7698 28.09.81 1250 1400 30
7698 BLAKE MANAGER 7839 01.05.81 2850 30
7782 CLARK MANAGER 7839 09.06.81 2450 10
7788 SCOTT ANALYST 7566 09.12.82 3000 20
7839 KING PRESIDENT 17.11.81 10000 10
7844 TURNER SALESMAN 7698 08.09.81 1500 0 30
7876 ADAMS CLERK 7788 12.01.83 1100 20
7900 JAMES CLERK 7698 03.12.81 950 30
7902 FORD ANALYST 7566 03.12.81 3000 20
7934 MILLER CLERK 7782 23.01.82 1300 10
14 rows selected.
Your code would then be like this; note line #6 which calculates row number as I suggested; you'll use it in the final WHERE clause (line #11), while presenting that "max" row number value you desperately wanted.
SQL> with temp as
2 (select deptno,
3 sal,
4 row_number() over (partition by deptno order by sal) rn,
5 --
6 row_number() over (partition by deptno order by sal desc) rnd
7 from emp
8 )
9 select deptno, sal, rn
10 from temp
11 where rnd = 1
12 /
DEPTNO SAL RN
---------- ---------- ----------
10 10000 3
20 3000 4
30 2850 6
SQL>
I want to select the top 5 rows and additional a 6th row named Others with rest aggregated.
with
Anzahl as
(SELECT
NVL (parse_listener_log_line (connect_string, 'HOST'), 'n/a') host, COUNT(*) cnt
FROM listener_log
WHERE ID_MANDANT = :P100_MANDANT
AND ID_SERVER = :P100_SERVER
GROUP BY parse_listener_log_line (connect_string, 'HOST')
ORDER BY cnt DESC),
client as
(select
case
when rownum > 4 then 'Others'
else host
end as client, cnt
FROM Anzahl)
SELECT client, cnt
FROM client;
CLIENT CNT
jdbc 118553
server2 106170
server1 101710
server4 13370
Others 8734
Others 1760
Others 1365
Others 1058
A little bit of analytic functions (row_number) along with set operations (union) might do what you're looking for.
Scott's EMP table contains these data:
SQL> select ename, sal from emp order by sal desc;
ENAME SAL
---------- ----------
KING 5000
FORD 3000
SCOTT 3000
JONES 2975
BLAKE 2850
CLARK 2450
ALLEN 1600
TURNER 1500
MILLER 1300
WARD 1250
MARTIN 1250
ADAMS 1100
JAMES 950
SMITH 800
14 rows selected.
Now: find each row's ordinal number (using row_number) and union the first 5 rows (take each of them as is) with the sixth one that contains aggregated salaries:
SQL> with temp as
2 (select ename,
3 sal,
4 row_number() over (order by sal desc) rn
5 from emp
6 )
7 select rn,
8 ename,
9 sal
10 from temp
11 where rn <= 5
12 union all
13 select 6,
14 'Other',
15 sum(sal)
16 from temp
17 where rn > 5
18 order by rn;
RN ENAME SAL
---------- ---------- ----------
1 KING 5000
2 SCOTT 3000
3 FORD 3000
4 JONES 2975
5 BLAKE 2850
6 Other 12200
6 rows selected.
SQL>
I have table : tb_user.
id|name|value
10|boy|500
20|Ony|200
10|boy|500
When I execute the following query:
Select id,name,sum(value) as grant_total from tb_user group by id,name
the result is:
id|name|grant_total
10|boy |1000
20|Ony |200
I want to add 1 column --> ranking
id|name|grant_total|ranking
10|boy |1000 |1
20|Ony |200 |2
how to make ranking?
Try:
Select id,name,sum(value) as grant_total, rownum as ranking from tb_user group by id,name
You could use ROWNUM and a subquery to first order the rows:
SQL> WITH DATA AS(
2 SELECT 10 ID, 'boy' NAME, 1000 grand_total FROM dual UNION ALL
3 SELECT 20, 'Ony', 200 grand_total from dual
4 )
5 SELECT t.*,
6 rownum ranking
7 FROM
8 ( SELECT * FROM DATA ORDER BY grand_total DESC
9 ) t
10 /
ID NAM GRAND_TOTAL RANKING
---------- --- ----------- ----------
10 boy 1000 1
20 Ony 200 2
SQL>
Or,
You could use the Analytic functions:
For example,
SQL> WITH DATA AS(
2 SELECT 10 ID, 'boy' NAME, 1000 grand_total FROM dual UNION ALL
3 SELECT 20, 'Ony', 200 grand_total from dual
4 )
5 SELECT t.*, row_number() OVER(ORDER BY grand_total DESC) ranking FROM DATA t;
ID NAM GRAND_TOTAL RANKING
---------- --- ----------- ----------
10 boy 1000 1
20 Ony 200 2
SQL>
Depending on the requirement, you need to use:
ROW_NUMBER
RANK
DENSE_RANK
Let's see an example,
SQL> SELECT empno, ename, sal FROM emp;
EMPNO ENAME SAL
---------- ---------- ----------
7369 SMITH 800
7499 ALLEN 1600
7521 WARD 1250
7566 JONES 2975
7654 MARTIN 1250
7698 BLAKE 2850
7782 CLARK 2450
7788 SCOTT 3000
7839 KING 5000
7844 TURNER 1500
7876 ADAMS 1100
7900 JAMES 950
7902 FORD 3000
7934 MILLER 1300
14 rows selected.
SQL> SELECT empno,
2 ename,
3 sal,
4 ROW_NUMBER() OVER(ORDER BY sal) rnum,
5 RANK() OVER(ORDER BY sal DESC) rank,
6 DENSE_RANK() OVER(ORDER BY sal) drank
7 FROM emp
8 ORDER BY empno
9 /
EMPNO ENAME SAL RNUM RANK DRANK
---------- ---------- ---------- ---------- ---------- ----------
7369 SMITH 800 1 14 1
7499 ALLEN 1600 8 7 7
7521 WARD 1250 4 10 4
7566 JONES 2975 11 4 10
7654 MARTIN 1250 5 10 4
7698 BLAKE 2850 10 5 9
7782 CLARK 2450 9 6 8
7788 SCOTT 3000 12 2 11
7839 KING 5000 14 1 12
7844 TURNER 1500 7 8 6
7876 ADAMS 1100 3 12 3
7900 JAMES 950 2 13 2
7902 FORD 3000 13 2 11
7934 MILLER 1300 6 9 5
14 rows selected.
SQL>
The easiest way is to use the ROW_NUMBER() analytic function (or RANK() if you want to rank ties equally):
SELECT id, name, SUM(value) AS grand_total
, ROW_NUMBER() OVER ( ORDER BY SUM(value) DESC ) AS ranking
FROM tb_user
GROUP BY id, name
ORDER BY ranking;
is there any way to write query with following functionality, add where clause as a conditional way,
select e.emp_id, emp.admin_user from employees e
if emp.admin != 'Y'
then
query run with where clause
else
query run without where clause ?
Using a CASE expression in the WHERE clause should do the trick. When you say you don't need the where clause if condition is not met, then all you want is a condition like WHERE 1 = 1, i.e. when condition is not met then return all rows. So, you need to make the not met condition as always TRUE.
For example,
I have an employee table,
SQL> SELECT empno, ename, deptno
2 FROM emp;
EMPNO ENAME DEPTNO
---------- ---------- ----------
7369 SMITH 20
7499 ALLEN 30
7521 WARD 30
7566 JONES 20
7654 MARTIN 30
7698 BLAKE 30
7782 CLARK 10
7788 SCOTT 20
7839 KING 10
7844 TURNER 30
7876 ADAMS 20
7900 JAMES 30
7902 FORD 20
7934 MILLER 10
14 rows selected.
SQL>
I want to select the employee details, if department is 20 then use the where clause else return all the employee details, but filter the department which meets the where condition.
SQL> SELECT empno, ename, deptno
2 FROM emp
3 WHERE ename =
4 CASE
5 WHEN deptno = 20
6 THEN 'SCOTT'
7 ELSE ename
8 END
9 /
EMPNO ENAME DEPTNO
---------- ---------- ----------
7499 ALLEN 30
7521 WARD 30
7654 MARTIN 30
7698 BLAKE 30
7782 CLARK 10
7788 SCOTT 20
7839 KING 10
7844 TURNER 30
7900 JAMES 30
7934 MILLER 10
10 rows selected.
SQL>
So, for department 20, the filter is applied by where clause, and I get only the row for ename SCOTT, for others it returns all the rows.
To keep it simple I would go for union clause in this case, so you can have your where clause as complex as you need. I tried to guess your table structure from above comment, let's see this example:
SQL> create table employees (emp_id number, admin_user number, project_id number);
Table created.
SQL> create table project_accessible_to_user (emp_id number, project_id number);
Table created.
Now make simple union all of two queries one with where condition anoother without it
SQL> select * from employees e where e.admin_user!='Y' and project_id in
(select project_id from project_accessible_to_user where emp_id=e.emp_id)
union all
select * from employees e where (e.admin_user is null or
e.admin_user='Y');
UNION ALL is better from performance point of view as UNION because it means that it is not checking for intersect values so if there are any it will return duplicates. However in this case it is filtered already by condition on admin_user, so these duplicates will not occure.
I want to display the column c.officeID along with the column "Amount". This is my query:
select c.officeID,max(Sum(p.amount)) as “Amount” from payment p, client c where c.clientid in (select clientid from client) and p.clientID=c.clientID group by c.officeID;
I tried using:
select c.officeID,max(Sum(p.amount)) as “Amount” from payment p, client c where c.clientid in (select clientid from client) and p.clientID=c.clientID group by c.officeID,p.amount;
But I am getting a error saying 'ORA00937-Not a single group-group funtion'. Could anyone please tell me where I am going wrong?
You can't add a max and group by, use this following query to get the result, if you want max then use the second query,
SCOTT#research 17-APR-15> select c.empno,Sum(p.sal) as "Amount"
2 from empp p, emp c
3 where c.empno in (select empno from emp)
4 and p.empno = c.empno
5 group by c.empno
6 ;
EMPNO Amount
---------- ----------
7782 2450
7839 5000
7844 1500
7698 2850
7521 1250
7902 3000
7566 2975
7654 1250
7788 3000
7934 1300
7499 1600
7876 1100
234 800
7900 950
14 rows selected.
select max("Amount") from (
select c.empno,Sum(p.sal) as "Amount"
from empp p, emp c
where c.empno in (select empno from emp)
and p.empno = c.empno
group by c.empno)
MAX("AMOUNT")
-------------
5000