Need help with Group functions
I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obviousI work with diagrams of the sample (in particular with the employees table):
DESC employees;
result
What I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:
SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
Employees emp employees JOIN Bishop
ON emp.manager_id = mgr.employee_id
GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
ORDER BY mgr.department_id;
result
As you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
All advice?
Hello
Welcome to the forum!
user12107811 wrote:
I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obvious
Just the opposite! Looks like a very difficult mission.
I work with diagrams of the sample (in particular with the employees table):
DESC employees;
resultWhat I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:
SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
Employees emp employees JOIN Bishop
ON emp.manager_id = mgr.employee_id
GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
ORDER BY mgr.department_id;
resultAs you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
All advice?
Yes, you're almost done. You just need to add one more condition. You have to calculate the average value of total_cnt (the COUNT (*) you already do) of a Department and compare that to total_cnt.
There are several ways to do this, including
a scalar subquery (in a HAVING clause)
(b) make a result set with one line per Department, containing the average_cnt and reach than your current result set
(c) analytical functions. Analytical functions are calculated after the GROUP BY clause is applied and aggregate functions are calculated, it is legitimate to say "AVG (COUNT (*)) MORE (...)").
If thinking (c) is the simplest. It involves the use of a query of Tahina, but (a) and (b) also require subqueries.
This sounds like homework, so I'll do it for you.
Instead, here is a very similar problem with the hr.employees table.
Let's say that we are interested in total wages given each type of work in each Department.
SELECT department_id
, job_id
, SUM (salary) AS sum_sal
FROM hr.employees
GROUP BY department_id
, job_id
ORDER BY department_id
, job_id
;
Results:
DEPARTMENT_ID JOB_ID SUM_SAL
------------- ---------- ----------
10 AD_ASST 4400
20 MK_MAN 13000
20 MK_REP 6000
30 PU_CLERK 13900
30 PU_MAN 11000
40 HR_REP 6500
50 SH_CLERK 64300
50 ST_CLERK 55700
50 ST_MAN 36400
60 IT_PROG 28800
70 PR_REP 10000
80 SA_MAN 61000
80 SA_REP 243500
90 AD_PRES 24000
90 AD_VP 34000
100 FI_ACCOUNT 39600
100 FI_MGR 12000
110 AC_ACCOUNT 8300
110 AC_MGR 12000
SA_REP 7000
Now suppose we want to find out which of these sum_sals is higher than the average sum_sal of his Department.
For example, in detriment 110 (near the end OIF the list) there two types of work (AC_ACCOUND and AC_MGR) that have sum_sals of 8300 and 12000. The average of these two numbers is 10150, so we selected AC_MGR (because its sum_sal, 12000, is superior to 10150, and we do not want to include AC_ACCOUNT, because its sum_sal, 8300, is less than or equal to the average of the Department.
In departments where there is only one job type (for example, Department 70, or null "Department" at the end of the list above) the only sum_sal will be the average; and because the sum_sal is not greater than the average, we want to exclude this line.
Let's start with the calculation of the avg_sum_sal using the analytical function AVG:
SELECT department_id
, job_id
, SUM (salary) AS sum_sal
, AVG (SUM (salary)) OVER (PARTITION BY department_id) AS avg_sum_sal
FROM hr.employees
GROUP BY department_id
, job_id
ORDER BY department_id
, job_id
;
Output:
DEPARTMENT_ID JOB_ID SUM_SAL AVG_SUM_SAL
------------- ---------- ---------- -----------
10 AD_ASST 4400 4400
20 MK_MAN 13000 9500
20 MK_REP 6000 9500
30 PU_CLERK 13900 12450
30 PU_MAN 11000 12450
40 HR_REP 6500 6500
50 SH_CLERK 64300 52133.3333
50 ST_CLERK 55700 52133.3333
50 ST_MAN 36400 52133.3333
60 IT_PROG 28800 28800
70 PR_REP 10000 10000
80 SA_MAN 61000 152250
80 SA_REP 243500 152250
90 AD_PRES 24000 29000
90 AD_VP 34000 29000
100 FI_ACCOUNT 39600 25800
100 FI_MGR 12000 25800
110 AC_ACCOUNT 8300 10150
110 AC_MGR 12000 10150
SA_REP 7000 7000
Now all we have to do is to compare the sum_sal and avg_sum_sal columns.
Given that the analytic functions are calculated after the WHERE clause is applied, we cannot use avg_sum_sal in the WHERE clause of the query, even where it has been calculated. But we can do that in a subquery; Then, we can use avg_sum_sal in any way that we love in the Super-requete:
WITH got_avg_sum_sal AS
(
SELECT department_id
, job_id
, SUM (salary) AS sum_sal
, AVG (SUM (salary)) OVER (PARTITION BY department_id) AS avg_sum_sal
FROM hr.employees
GROUP BY department_id
, job_id
)
SELECT department_id
, job_id
, sum_sal
FROM got_avg_sum_sal
WHERE sum_sal > avg_sum_sal
ORDER BY department_id
, job_id
;
Results:
DEPARTMENT_ID JOB_ID SUM_SAL
------------- ---------- ----------
20 MK_MAN 13000
30 PU_CLERK 13900
50 SH_CLERK 64300
50 ST_CLERK 55700
80 SA_REP 243500
90 AD_VP 34000
100 FI_ACCOUNT 39600
110 AC_MGR 12000
Tags: Database
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t
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---------- ---------- --- ---------- ----------
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SQL >
SY.
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I need the installation:
user names, authentication group etc.
Thank you!
Take a peek inside has the below examples of config - everything you need: -.
http://www.Cisco.com/en/us/products/ps5854/prod_configuration_examples_list.html
HTH >
Andrew.
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