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0 From time to time there is a need in my SQL queries to have access to a table with a specific number of records. Sometimes I want to explode the results from a query – double, treble etc. the number of records. On other occasions I want to select from a specific range or outer join with a specified range to have an entry for every rank, date etc. More concrete examples (in terms of EMP and DEPT):
- select the number of employees hired on every day of the week – include entries for days on which no one was hired
- select the employees ranked on even positions when ordered by salary (note: there are many ways to approach this)
- present for the numbers 1 to 5 the number of employees that manage that number of subordinates
- select all employees as many times as there are Clerks
Of course the old way of doing this was create a utility or helper table that contained a very large number of records (something like dual but with many rows). Each row typically only contained a number. At other times you also have created tables with seven records (days of the week), twelve records (months of the year), 26 records (letters in the alphabet) etc. However, these could all have been derived in views from the one table with just many numeric values. You could also fake this table by using a view or table in your schema that you know to be always available and contain a substantial number of records. Tom Kyte’s favorite for example is ALL_OBJECTS. We will look at the ‘modern’ way: TABLE FUNCTIONS (Oracle 9i) and the MODEL clause (10g).
Using Table-Functions
Much has been written about Table Functions that I will not repeat here. Basically, a Table Function is a PL/SQL function that returns a Nested Table. Such a function can be used in SQL queries, with the function wrapped inside a TABLE( ) operator. We can use TABLE(function) just like any table, view or in-line view in the query; it is just another result-set or record-source.
A useful, generic Table-Function is the following:
create type number_table as table of number
create or replace function number_tbl
( p_lower in number:=0, p_upper in number:=10)
return number_table
is
l_num_tbl number_table := number_table();
begin
for i in p_lower..p_upper loop
l_num_tbl.extend;
l_num_tbl(l_num_tbl.last):= i;
end loop;
return l_num_tbl;
end number_tbl;
Using this function number_tbl we have a row-generator at our disposal that can be used in many ways.
List the days of the week (knowing that 21st november 2004 is a Sunday
SELECT to_char( days.column_value + to_date('21-11-2004','DD-MM-YYYY'),'DAY') day
FROM table(number_tbl(0,6)) daysThe result:
DAY --------- SUNDAY MONDAY TUESDAY WEDNESDAY THURSDAY FRIDAY SATURDAY
Find for the numbers 0 through 5 the number of managers who have that number of subordinates:
SELECT num_subs.column_value number_of_subordinates
, count(subs_count.cnt) number_of_mgrs
FROM table(number_tbl(0,5)) num_subs
left outer join
(select count(*) cnt from emp group by mgr) subs_count
on num_subs.column_value = subs_count.cnt
group
by num_subs.column_valueResults:
NUMBER_OF_SUBORDINATES NUMBER_OF_MGRS
---------------------- --------------
0 0
1 4
2 1
3 1
4 0
5 1Select the employees ranked 5 through 7 when ordered by salary, descending:
select *
from table( number_tbl(5, 7)) num
, ( select emp.*, rownum rn
from ( select emp.*
from emp
order
by sal desc) emp
) emp
where rn = num.column_valueCOLUMN_VALUE EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO RN
------------ ---------- ---------- --------- ---------- --------- ---------- ---------- ---------- -
5 7698 BLAKE MANAGER 7839 01-MAY-81 2850 30 5
6 7782 CLARK MARKETEER 7839 09-JUN-81 2450 10 6
7 7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30 7
Using the 10g Model Clause
Inspired by Tom Kyte’s column in Oracle Magazine of November 2004, I realized how you can make use of the MODEL clause in Oracle 10g to have records appear out of thin air. The advantage of this approach is that we do not have to create TYPEs and FUNCTIONs before we can start querying.
To select the number of employees hired in the years 1979 through 1983:
select years.year
, count(empno)
from ( SELECT 1979+cell year
FROM DUAL
MODEL
DIMENSION BY (0 attr)
MEASURES (0 cell)
RULES ITERATE (5)
( cell[iteration_number] = iteration_number)
) years
left outer join
emp
on (extract(year from hiredate) = years.year)
group
by year
ORDER
BY year
The outcome, in case you were wondering:
YEAR COUNT(EMPNO) ------ ------------ 1979 0 1980 1 1981 10 1982 1 1983 0
The in-line view years makes use of DUAL and explodes the set returned from dual. It builds a two-dimensional array:
SELECT cell , attr
FROM DUAL
MODEL
DIMENSION BY (0 attr)
MEASURES (0 cell)
RULES ITERATE (5)
( cell[iteration_number] = iteration_number)
CELL ATTR
----- ----------
0 0
1 1
2 2
3 3
4 4
With the ITERATE rule we specify the number of records that should be created in th model. With the DIMENSION and MEASURES, we define the composition of each individual record. Finally we use the almost simplest assignment expression for the cell (MEASURE) values.
In a similar fashion we can get a list of the days of the week:
SELECT cell day_number
, to_char( cell + to_date('21-11-2004','DD-MM-YYYY'),'DAY') day
FROM DUAL
MODEL
DIMENSION BY (0 attr)
MEASURES (0 cell)
RULES ITERATE (7)
( cell[iteration_number] = iteration_number)
DAY_NUMBER DAY
---------- ---------
0 SUNDAY
1 MONDAY
2 TUESDAY
3 WEDNESDAY
4 THURSDAY
5 FRIDAY
6 SATURDAY
Here I happen to use a SUNDAY as starting point (21st November 2004, today).
Using the MODEL clause in this fashion, you can any number of records you desire with values that are either from 1 to the number of records or derived from those values – such as dates, ascii characters etc. Get a list of all letters in the alphabet:
SELECT chr(97+cell) letter FROM DUAL MODEL DIMENSION BY (0 attr) MEASURES (0 cell) RULES ITERATE (26) ( cell[iteration_number] = iteration_number) L - a b c d e f g h i j ...
Resources on the MODEL Clause:
SQL Reference (Oracle 10g On Line Documentation) – Model Expressions
SQL Reference (Oracle 10g On Line Documentation) – Model Clause
Data Warehousing Guide (Oracle 10g On Line Documentation) – 22. SQL For Modeling
Partition Outer Join (10g)
In certain circumstances you may also want to make use of the Partition Outer Join Clause. For example, if we want to select from EMP the number of employees hired in the years 1980, 1981 and 1982 per department – we need a partition outer join to ensure that we will have a result for each of these years for each department.
If we execute this query – to find the number of employees hired in each year for every department:
select years.year
, deptno
, count(empno)
from ( SELECT 1980+cell year
FROM DUAL
MODEL
DIMENSION BY (0 attr)
MEASURES (0 cell)
RULES ITERATE (3)
( cell[iteration_number] = iteration_number)
) years
left outer join
emp
on (extract(year from hiredate) = years.year)
group
by year
, deptno
ORDER
BY year
, deptno
the result is somewhat incomplete:
YEAR DEPTNO COUNT(EMPNO) ------ ---------- ------------ 1980 20 1 1981 10 1 1981 20 3 1981 30 6 1982 10 1
No entries for departments 10 and 30 in 1980 nor entries in 1982 for departments 20 and 30. With the partition outer join clause we can complement this result:
select years.year
, deptno
, count(empno)
from ( SELECT 1980+cell year
FROM DUAL
MODEL
DIMENSION BY (0 attr)
MEASURES (0 cell)
RULES ITERATE (3)
( cell[iteration_number] = iteration_number)
) years
left outer join
emp partition by (deptno)
on (extract(year from hiredate) = years.year)
group
by year
, deptno
ORDER
BY year
, deptno
Here we have “left outer joined” years with emp. That can be read as: take all left records and find joinable records from emp; if you can find none, add an “empty” record representing the right side – emp. In the previous query, since every year could be matched with at least one department’s score, there were no empty records added; hence we had no row at all for the departments 10 and 30 for 1980. In this query, we have added partition by (deptno) to emp. This tells the SQL Engine that the assignment to left outer join years to emp must be executed for every deptno-partition-of-emp. The set of records returned from EMP is partitioned by deptno (three partitions). Then, the outer join is performed again for each of the three partitions. This must results in a record for every year and every partition (i.e. deptno). The result of this query:
YEAR DEPTNO COUNT(EMPNO) ------ ---------- ------------ 1980 10 0 1980 20 1 1980 30 0 1981 10 1 1981 20 3 1981 30 6 1982 10 1 1982 20 0 1982 30 0
Related posts:
Poor man's VPD – Virtual Private Database before 8i and in Standard Edition Databases Oracle 10gR2 – Conditional PL/SQL Compilation 
Dynamic SQL Pivoting – Stealing Anton’s Thunder Putting Analytical Functions to good use – find tables with multiple foreign keys to the same referenced table Pimp you SQL Developer
what do you think about this?
select Job, DECODE(DEPTNO,10,sum(SAL), NULL) Dept 10,
DECODE(DEPTNO,20,sum(SAL), NULL) Dept 20,
DECODE(DEPTNO,30,sum(SAL), NULL) Dept 10,
sum(SAL) Total
from EMP
group by Job;
I am trying to create a view that shows the total of overdue people by, less than 30 days, 31 to 59days, and over 60 days overdue.
What is wrong with my code.
CREATE OR REPLACE VIEW csu_status_date_vu
(category, less than 30 days, 31-59 days, 60 days and greater)
AS SELECT ass_attribute9, effective_start_date
FROM PER_ALL_ASSIGNMENTS_F
WHERE (SELECT SUM (CASE
WHEN effective_start_date > TRUNC (SYSDATE – 30)
THEN 1
ELSE 0
END) “less than 30 days”,
SUM (CASE
WHEN effective_start_date = TRUNC (SYSDATE – 59)
THEN 1
ELSE 0
END) “31 -59 days”,
SUM (CASE
WHEN effective_start_date
Re the pivot query, isn’t this a whole lot simpler? :-
SQL> select job
2 , sum(case when deptno=10 then sal else 0 end) dept10
3 , sum(case when deptno=20 then sal else 0 end) dept20
4 , sum(case when deptno=30 then sal else 0 end) dept30
5 , sum(sal) total
6 from emp
7 group by job;
JOB DEPT10 DEPT20 DEPT30 TOTAL
——— ———- ———- ———- ———-
ANALYST 0 6000 0 6000
CLERK 1300 1900 950 4150
MANAGER 2450 2975 2850 8275
PRESIDENT 5000 0 0 5000
SALESMAN 0 0 5600 5600
OK, using the SQL MODEL clause (only in Oracle 10g I am afraid) I found the near perfect solution. Its result looks as follows:
JOB DEPT10 DEPT20 DEPT30 GRAND_TOTAL--------- ---------- ---------- ---------- -----------
CLERK 1300 1900 950 4150
SALESMAN 5600 5600
PRESIDENT 5000 5000
MANAGER 2450 2975 2850 8275
ANALYST 6000 6000
The query itself:
select job
, dept10
, dept20
, dept30
, grand_total
from (select distinct deptno, job, sum(sal) OVER ( PARTITION BY deptno, job) sumsal , sum(sal) over( partition by job) sumjob from emp )
model
return updated rows
partition by (job)
dimension by (deptno)
measures (sumsal, lpad(' ',10) dept10, lpad(' ',10) dept20, lpad(' ',10) dept30 , sumjob grand_total)
rules upsert
(
dept10 [0] = sumsal [10]
, dept20 [0] = sumsal [20]
, dept30 [0] = sumsal [30]
, grand_total [0] = max(grand_total) [ANY]
)
For more insight and an explanation, see this post Pivoting in SQL using the 10g Model Clause
OK, I missed the proper pivot. By replacing job and deptno in the above query we get the following result:
The query now looks as follows:
select job , substr(sal_sum1, 1, 20) sal_sum1 , substr(sal_sum2, 1, 20) sal_sum2 , substr(sal_sum3, 1, 20) sal_sum3 , substr(sal_sum4, 1, 20) sal_sum4 , substr(sal_sum5, 1, 20) sal_sum5 , substr(sal_sum6, 1, 20) sal_sum6 from ( select job , deptno||': '||sal_sum sal_sum1 , lead(deptno,1) over (partition by job order by deptno) ||': '||lead(sal_sum,1) over (partition by job order by deptno) sal_sum2 , lead(deptno,2) over (partition by job order by deptno) ||': '||lead(sal_sum,2) over (partition by job order by deptno) sal_sum3 , lead(deptno,3) over (partition by job order by deptno) ||': '||lead(sal_sum,3) over (partition by job order by deptno) sal_sum4 , lead(deptno,4) over (partition by job order by deptno) ||': '||lead(sal_sum,4) over (partition by job order by deptno) sal_sum5 , lead(deptno,5) over (partition by job order by deptno) ||': '||lead(sal_sum,5) over (partition by job order by deptno) sal_sum6 , row_number() over (partition by job order by deptno) rnk from ( select distinct job , deptnos.deptno , sum(sal) over (partition by emp.deptno, job) sal_sum from emp partition by (job) right outer join (select distinct deptno from emp) deptnos on (emp.deptno = deptnos.deptno) order by job , deptno ) ) where rnk = 1Maybe a better, more generic solution with less hard-coding and less coding in general – though only suitable for Oracle 10g because of the partion by clause in the outer join expression – would render the following result:
DEPTNO SAL_SUM1 SAL_SUM2 SAL_SUM3 SAL_SUM4 SAL_SUM5 SAL_SUM6 ------- -------------------- -------------------- -------------------- -------------------- ----- 10 ANALYST: CLERK: 1300 MANAGER: 2450 PRESIDENT: 5000 SALESMAN: : 20 ANALYST: 6000 CLERK: 1900 MANAGER: 2975 PRESIDENT: SALESMAN: : 30 ANALYST: CLERK: 950 MANAGER: 2850 PRESIDENT: SALESMAN: 5600 :The code used to produce this result:
select deptno , substr(sal_sum1, 1, 20) sal_sum1 , substr(sal_sum2, 1, 20) sal_sum2 , substr(sal_sum3, 1, 20) sal_sum3 , substr(sal_sum4, 1, 20) sal_sum4 , substr(sal_sum5, 1, 20) sal_sum5 , substr(sal_sum6, 1, 20) sal_sum6 from ( select deptno , job||': '||sal_sum sal_sum1 , lead(job,1) over (partition by deptno order by job) ||': '||lead(sal_sum,1) over (partition by deptno order by job) sal_sum2 , lead(job,2) over (partition by deptno order by job) ||': '||lead(sal_sum,2) over (partition by deptno order by job) sal_sum3 , lead(job,3) over (partition by deptno order by job) ||': '||lead(sal_sum,3) over (partition by deptno order by job) sal_sum4 , lead(job,4) over (partition by deptno order by job) ||': '||lead(sal_sum,4) over (partition by deptno order by job) sal_sum5 , lead(job,5) over (partition by deptno order by job) ||': '||lead(sal_sum,5) over (partition by deptno order by job) sal_sum6 , row_number() over (partition by deptno order by job) rnk from ( select distinct deptno , jobs.job , sum(sal) over (partition by emp.job, deptno) sal_sum from emp partition by (deptno) right outer join (select distinct job from emp) jobs on (emp.job = jobs.job) order by deptno , job ) ) where rnk = 1I am not sure this is the most elegant solution, but it most certainly gives the matrix report you asked for:
JOB DEPT10 DEPT20 DEPT30 TOTAL
-------- ---------- ---------- ---------- ----------
Clerks 1430 2090 1045 4565
Managers 0 2975 2850 5825
Salesmen 0 0 3100 3100
select case dn.n
when 1
then 'Clerks'
when 2
then 'Managers'
when 3
then 'Salesmen'
else null
end job
, max( case
when rn=1 and dn.n = 1
then clerks
when rn=1 and dn.n = 2
then managers
when rn=1 and dn.n = 3
then salesmen
else null
end) dept10
, max( case
when rn=2 and dn.n = 1
then clerks
when rn=2 and dn.n = 2
then managers
when rn=2 and dn.n = 3
then salesmen
else null
end ) dept20
, max(case
when rn=3 and dn.n = 1
then clerks
when rn=3 and dn.n = 2
then managers
when rn=3 and dn.n = 3
then salesmen
else null
end) dept30
, sum(case
when dn.n = 1
then clerks
when dn.n = 2
then managers
when dn.n = 3
then salesmen
else 0
end) total
from
(select deptno
, row_number() over (order by deptno) rn
, sum( case job
when 'CLERK'
then sal
else 0
end
) clerks
, sum( case job
when 'MANAGER'
then sal
else 0
end
) Managers
, sum( case job
when 'SALESMAN'
then sal
else 0
end
) salesmen
from emp
group
by deptno
)
, (select rownum n from emp where rownum<4) dn
group
by case dn.n
when 1
then 'Clerks'
when 2
then 'Managers'
when 3
then 'Salesmen'
else null
end
There are some good examples here but don’t forget that in many circumstances an alternative approach
for magicing rows into existence is to deliberately make a cartesian join to this type of in-line view:
select rownum
from {large table}
where rownum < {rows that you want}
how do i create a matrix query to display job, salary for that job based on dept no., and total salary for that job fro all departments,
job dept10 dept20 dept30 total
—- —— —— —— —–
clerk 1300 1900 950 4150
using grouping functions
how do i create a matrix query to display job, salary for that job based on dept no., and total salary for that job fro all departments,
job dept10 dept20 dept30 total
—- —— —— —— —–
clerk 1300 1900
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