CODEWITHSUNDEEP
sqloracle-database
Tom
Tom

Reputation: 12988

Oracle SQL - Set consecutive numbers as a range

This query returns the following results...

WITH t(wk, COST) AS
(SELECT wk, COST FROM myTable WHERE id = '345')
SELECT listagg(wk,',') WITHIN GROUP(ORDER BY wk) AS wks, COST 
FROM   t
GROUP BY COST;

...

WKS               COST
----------------------
17, 18, 19, 21    446
26, 27, 28        588

Is it possible to create the following results where consecutive weeks are returned as from and to. (eg 1-10 rather than 1, 2, 3, 4,... etc)

WKS               COST
----------------------
17-19, 21         446
26-28             588

Upvotes: 3

Views: 642

Answers (1)

Noel
Noel

Reputation: 10525

In this problem you have to identify the sequence of consecutive weeks and group them. Here's my solution.

  • Use the LAG function to identify any breaks in the sequence.
  • Use the SUM function to assign a group number to each sequence.
  • Find the starting and ending week in each group.
  • Finally use the LISTAGG function to aggregate the result.

Query:

with x(wk, cost, startgroup) as(
    --identify the start of a sequence
    select wk, cost,
           case when wk = lag(wk,1) over (partition by cost order by wk) + 1
                then 0
                else 1
             end
    from mytable
    where id = '345'
    ),
    y(wk, cost, grp) as(
    --assign group number
    select wk, cost,
           sum(startgroup) over (partition by cost order by wk)
    from x
    ),
    z(wk, cost, grp) as(
    --get the max/min week for each group
    select case when min(wk) = max(wk)
                then cast(min(wk) as varchar2(10))
                else min(wk) ||'-'||max(wk)
                end,
            cost, grp
    from y
    group by cost, grp
    )
--aggregate by cost
select listagg(wk,',') within group(order by grp),
cost
from z
group by cost;

Demo at sqlfiddle.

Upvotes: 3

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