Oracle求值的連續範圍

首先帶上原文的鏈結：

有資料如下：

求v_name的連續區間。

期望的結果：

容易理解的寫法:

select v_name,

decode(count(*), 1, to_char(min(id)), min(id) || '-' || max(id)) b
from (select id, v_name, max(rn) over(partition by v_name order by id) rn
from (select id,
v_name,
decode(id - lag(id)
over(partition by v_name order by id),
1,0,
id) rn
from tmp_t))
group by v_name, rn
order by v_name, rn;

select v_name,

decode(to_char(mi),to_char(ma) ,to_char(ma),to_char(mi) || '-' || to_char(ma)) b
from (select v_name, min(id) mi, max(id) ma
from (select t.* from tmp_t t order by t.v_name, t.id)
group by v_name, id - rownum
order by v_name);

--連續id減行號的差值一定是相同的

select v_name,
decode(count(*), 1, to_char(min(id)), min(id) || '-' || max(id)) b
from (select t.*, id - row_number() over(partition by v_name order by id) gp
from tmp_t t)
group by v_name, gp
order by v_name

with tmp_t as(

select to_date('201401','yyyy-mm') as v_month,100 as v_value from dual union all
select to_date('201402','yyyy-mm'),100 from dual union all
select to_date('201404','yyyy-mm'),100 from dual union all
select to_date('201405','yyyy-mm'),300 from dual union all
select to_date('201406','yyyy-mm'),100 from dual union all
select to_date('201311','yyyy-mm'),110 from dual union all
select to_date('201310','yyyy-mm'),110 from dual union all
select to_date('201407','yyyy-mm'),100 from dual )
select v_value,decode(to_char(min(v_month), 'yyyymm'),
to_char(max(v_month), 'yyyymm'),
to_char(max(v_month), 'yyyymm'),
to_char(min(v_month), 'yyyymm') || '-' ||
to_char(max(v_month), 'yyyymm')) v_range
from (select v_month,
v_value,
to_number(to_char(add_months(v_month,
-1 * row_number()
over(partition by v_value order by
v_month)),
'yyyymm')) diff
from tmp_t)
group by diff, v_value
order by 1;

全文完。

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