Oracle SQL-将连续数字设置为一个范围

此查询返回以下结果

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;

是否可以创建以下结果，其中连续的周返回为from和to。（例如1-10而不是1、2、3、4等）

---

在这个问题中，您必须确定连续几周的顺序并将其分组。这是我的解决办法

• 使用滞后功能识别序列中的任何中断
• 使用SUM函数为每个序列分配一个组号
• 找出每组的开始和结束周
• 最后，使用listag函数聚合结果

查询：

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;

---

演示在。

第二行不是应该是“26-28”吗？很好！我会改变的…我需要一段时间来思考这个问题，但这正是我想要的。。。谢谢

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

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;