Get collection of integers from counter - sql

Question

Which would be the best way to get the number of people hired on each day of the week for 7 years from a table People that has their entry_date with a day-month-year as 01-Jun-91.
For example:
2000 2001 2002 etc..
SUN 2 0 1
MON 0 0 2
Do I have to create a counter for each day of each year? Like Sun2000, Sun2001 etc?

Answer 1

You need to join each day of the week with your entry_date and pivot the results.
SQL Fiddle
Query:
with x(days) as (
select 'sunday' from dual union all
select 'monday' from dual union all
select 'tuesday' from dual union all
select 'wednesday' from dual union all
select 'thursday' from dual union all
select 'friday' from dual union all
select 'saturday' from dual
)
select * from (
select x.days,
extract(year from emp.entry_date) entry_year
from x left outer join emp
on x.days = to_char(emp.entry_date,'fmday')
)
pivot(count(entry_year)
for entry_year in (
2007,
2008,
2009,
2010,
2011,
2012
)
)
order by
case days when 'sunday' then 1
when'monday' then 2
when'tuesday' then 3
when'wednesday' then 4
when'thursday' then 5
when'friday' then 6
when'saturday' then 7
end
Results:
| DAYS | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 |
|-----------|------|------|------|------|------|------|
| sunday | 0 | 0 | 0 | 0 | 0 | 0 |
| monday | 0 | 0 | 0 | 2 | 0 | 0 |
| tuesday | 0 | 0 | 0 | 0 | 1 | 0 |
| wednesday | 0 | 0 | 0 | 1 | 2 | 1 |
| thursday | 0 | 0 | 0 | 0 | 0 | 3 |
| friday | 0 | 0 | 0 | 0 | 0 | 0 |
| saturday | 0 | 0 | 0 | 0 | 0 | 0 |

Answer 2

You need to use group by on the year and entry_date to get the count of employees joined for each date.
For example:
Rem -- Assuming following table structure
create table People(id number, name varchar2(20), entry_date date);
Rem -- Following groups the results
select extract(year from entry_date) "Year", entry_date, count(id)
from People
where extract(year from entry_date) between 2008 and 2015
group by extract(year from entry_date), entry_date
order by extract(year from entry_date), entry_date;
Check this sqlfiddle to explore more.

Answer 3

Depending on the version of Oracle you're using (10g doesn't have the PIVOT function, for example), you might try something like the following conditional aggregation:
SELECT day_abbrev
, SUM(CASE WHEN year_num = 2000 THEN person_cnt ELSE 0 END) AS "2000"
, SUM(CASE WHEN year_num = 2001 THEN person_cnt ELSE 0 END) AS "2001"
, SUM(CASE WHEN year_num = 2002 THEN person_cnt ELSE 0 END) AS "2002"
, SUM(CASE WHEN year_num = 2003 THEN person_cnt ELSE 0 END) AS "2003"
, SUM(CASE WHEN year_num = 2004 THEN person_cnt ELSE 0 END) AS "2004"
, SUM(CASE WHEN year_num = 2005 THEN person_cnt ELSE 0 END) AS "2005"
, SUM(CASE WHEN year_num = 2006 THEN person_cnt ELSE 0 END) AS "2006"
FROM (
SELECT TO_CHAR(entry_date, 'DY') AS day_abbrev
, EXTRACT(YEAR FROM entry_date) AS year_num
, COUNT(*) AS person_cnt
FROM people
GROUP BY TO_CHAR(entry_date, 'DY'), EXTRACT(YEAR FROM entry_date)
) GROUP BY day_abbrev
ORDER BY TO_CHAR(NEXT_DAY(SYSDATE, day_abbrev), 'D');