Question

假设有如下记录：

Employee_id, work_start_date, work_end_date

1, 01-jan-2014, 07-jan-2014
1, 03-jan-2014, 12-jan-2014
1, 23-jan-2014, 25-jan-2014
2, 15-jan-2014, 25-jan-2014
2, 07-jan-2014, 15-jan-2014
2, 09-jan-2014, 12-jan-2014

要求是编写一个SQL select语句，该语句将汇总按employee_id分组的工作日，但排除重叠的句点（意思是 - 只将它们计算一次）。

所需的输出是：

Employee_id, worked_days

1, 13
2, 18

日期范围内的工作日计算如下：如果work_start_date = 5且work_end_date = 9则则working_days = 4（9 - 5）。

我可以编写一个pl / sql函数来解决这个问题（手动迭代记录并进行计算），但我确信可以使用SQL来实现更好的性能。

有人可以指出我正确的方向吗？

谢谢！

Answer 1

这是一个类似问题的略微修改的查询：
compute sum of values associated with overlapping date ranges

SELECT "Employee_id",
       SUM( "work_end_date" - "work_start_date" )
FROM(
  SELECT "Employee_id",
         "work_start_date" ,
         lead( "work_start_date" ) 
             over (Partition by "Employee_id"
                  Order by "Employee_id", "work_start_date" ) 
         As "work_end_date"
  FROM (
     SELECT "Employee_id", "work_start_date"
     FROM Table1
     UNION
     SELECT "Employee_id","work_end_date"
     FROM Table1
  ) x
) x
WHERE EXISTS (
   SELECT 1 FROM Table1 t
   WHERE t."work_start_date" > x."work_end_date"
     AND t."work_end_date" > x."work_start_date"
      OR t."work_start_date" = x."work_start_date"
     AND t."work_end_date" =  x."work_end_date"
)
GROUP BY "Employee_id"
;

演示：http://sqlfiddle.com/#!4/4fcce/2

Answer 2

这是一个棘手的问题。例如，您无法使用lag()，因为重叠时段可能不是＆＃34;之前的＆＃34;一。或者不同的时期可以在同一天开始或停止。

这个想法是重建时期。这该怎么做？查找期间开始的记录 - 也就是说，与其他任何记录都没有重叠。然后将其用作标志并累计计算此标志以计算重叠组。然后获得工作日只是从那里汇总：

with ps as (
      select e.*,
             (case when exists (select 1
                                from emps e2
                                where e2.employee_id = e.employee_id and
                                      e2.work_start_date <= e.work_start_date and
                                      e2.work_end_date >= e.work_end_date
                         )
                   then 0 else 1
            ) as IsPeriodStart
      from emps e
     )
select employee_id, sum(work_end_date - work_start_date) as Days_Worked
from (select employee_id, min(work_start_date) as work_start_date,
             max(work_end_date) as work_end_date
      from (select ps.*,
                   sum(IsPeriod_Start) over (partition by employee_id
                                             order by work_start_date
                                            ) as grp
            from ps 
           ) ps
      group by employee_id, grp
     ) ps
group by employee_id;

Answer 3

date_tbl type

create or replace package RG_TYPE is
  type date_tbl is table of date;
end;

函数（结果为具有2个参数之间的日期的表）

create or replace function dates
(
    p_from date,
    p_to date
) return rg_type.date_tbl pipelined
is
  l_idx date:=p_from;
begin
  loop
    if l_idx>nvl(p_to,p_from) then
      exit;
    end if;
    pipe row(l_idx);
    l_idx:=l_idx+1;
  end loop;
  return;
end;

SQL：

select employee_id,sum(c)
from
  (select e.employee_id,d.column_value,count(distinct w.employee_id) as c
  from   (select distinct employee_id from works) e,
         table(dates((select min(work_start_date) as a from works),(select max(work_end_date) as b from works))) d,
         works w
  where e.employee_id=w.employee_id
        and d.column_value>=w.work_start_date
        and d.column_value<w.work_end_date
  group by e.employee_id,d.column_value) Sub
group by employee_id  
order by 1,2

日期的工作日总和范围从多个记录（重叠）

3 个答案: