Question

这将是从表中获得每周每天雇用7年的人数的最佳方式。这些人的入职日期为每月 - 月 - 年，为01-Jun-91。

例如：

     2000 2001 2002 etc..

SUN   2    0    1
MON   0    0    2

我是否必须为每年的每一天创建一个柜台？像Sun2000，Sun2001等？

Answer 1

您需要使用entry_date和pivot结果加入一周中的每一天。

SQL Fiddle

<强>查询：

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

<强> 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

您需要使用group by和year上的entry_date来获取每个日期加入的员工数量。

例如：

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;

检查this sqlfiddle以了解详情。

Answer 3

根据您使用的Oracle版本（例如，10g没有PIVOT功能），您可以尝试以下条件聚合：

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');

从计数器获取整数的集合

3 个答案: