正如gelonsoft 所指出的，有很多方法可以做到这一点。我宁愿只打一次桌子。这是我目前最喜欢的，基于我第一次在这个网站上找到的方法，可能是在这个答案上（你可能会在这些问题中找到更多的方法：
select inventory_group_id, code, start_num,
case when end_num = start_num then null else end_num end as end_num
from (
select inventory_group_id, code, min(num) as start_num, max(num) as end_num
from (
select inventory_group_id, num, code,
row_number() over (order by num)
- row_number() over (partition by code order by num) as chain
from inventory_num
)
group by inventory_group_id, code, chain
)
order by 1,3;
INVENTORY_GROUP_ID C START_NUM END_NUM
------------------ - ---------- ----------
100003894 E 211 213
100003894 I 214
100003894 E 515 519
100003894 C 520 521
100003894 E 522
100003894 I 523
100003894 E 524 527
内部选择通过基于code和num值创建人工分组来完成所有工作 - 自行运行以查看它在做什么。下一个查询是折叠组以找到num每个人工组的最低和最高。最后的外部查询纯粹是为了如果链只有一个链接 - 即和是相同的 - 你提到你想要的end值。nullstartend
你没有说的是这些num值是否必须是连续的。如果我用code='I'and添加一条记录num=216，我得到相同数量的输出，但214-216被视为一个链，即使215两者之间没有值：
INVENTORY_GROUP_ID C START_NUM END_NUM
------------------ - ---------- ----------
100003894 E 211 213
100003894 I 214 216
...
我还没有想出如何调整这种row_number()方法以考虑到这一点并将它们视为单独的链 - 我很想看看它是否可以在保持简单的同时完成。给出的其他答案也会发生同样的事情；但我不确定这对你是否重要。
如果是这样，这是另一个版本，它只上桌一次；相当复杂，并且在这种形式下，与非连续num值具有相同的潜在问题：
select distinct inventory_group_id, code,
case
when prev_code = code then lag(num) over (order by rn)
else num
end as start_num,
case
when next_code != code and prev_code != code then null
when next_code = code then lead(num) over (order by rn)
else num
end as end_num
from (
select inventory_group_id, num, code, prev_code, next_code,
rownum as rn
from (
select inventory_group_id, num, code,
lag(code) over (partition by inventory_group_id
order by num) as prev_code,
lead(code) over (partition by inventory_group_id
order by num) as next_code
from inventory_num
)
where (prev_code is null or code != prev_code)
or (next_code is null or code != next_code)
order by 1,2,3
)
order by 1,3,2;
INVENTORY_GROUP_ID C START_NUM END_NUM
------------------ - ---------- ----------
100003894 E 211 213
100003894 I 214
100003894 E 515 519
100003894 C 520 521
100003894 E 522
100003894 I 523
100003894 E 524 527
内部查询从表中进行选择，并使用lead和lag 分析函数在每一行的任一侧查找代码。
下一个查询排除与下一行和上一行具有相同代码的任何行；也就是说，任何位于不间断链中间的东西。这意味着每个链最多折叠到两行，其中包含num该链的开始值和结束值。
外部查询中的语句通过使用和再次case使每个链的两行看起来相同；如果只有一行（例如 for ），则第二行的第一个子句构成最终值，你说你想要的。然后删除创建的重复项。leadlag214whencasenulldistinctcase
我建议您分别运行每个级别的查询，以查看它在做什么，并了解它对上一个级别的操作。
code='I'正如我所说，如果我用and引入一行，这会产生同样的潜在问题num=216：
INVENTORY_GROUP_ID C START_NUM END_NUM
------------------ - ---------- ----------
100003894 E 211 213
100003894 I 214 216
...
通过采用这种方法可以将其分成两个链，但它有点复杂，因为您必须跟踪和比较num值以及code值：
select distinct inventory_group_id, code,
case
when prev_num is null then num
when prev_code = code then lag(num) over (order by rn)
else num
end as start_num,
case
when next_code != code and prev_code != code then null
when next_code is null then num
when next_num is null then null
when next_code = code then lead(num) over (order by rn)
else num
end as end_num
from (
select inventory_group_id, num, code, prev_code, next_code,
case
when prev_num != num - 1 then null
else prev_num
end as prev_num,
case
when next_num != num + 1 then null
else next_num
end as next_num,
rownum as rn
from (
select inventory_group_id, num, code,
lag(code) over (partition by inventory_group_id
order by num) as prev_code,
lead(code) over (partition by inventory_group_id
order by num) as next_code,
lag(num) over (partition by inventory_group_id
order by num) as prev_num,
lead(num) over (partition by inventory_group_id
order by num) as next_num
from inventory_num
)
where (prev_code is null or code != prev_code)
or (next_code is null or code != next_code)
or (prev_num is null or num != prev_num + 1)
or (next_num is null or num != next_num - 1)
order by 1,2,3
)
order by 1,3,2;
INVENTORY_GROUP_ID C START_NUM END_NUM
------------------ - ---------- ----------
100003894 E 211 213
100003894 I 214
100003894 I 216
100003894 E 515 519
100003894 C 520 521
100003894 E 522
100003894 I 523
100003894 E 524 527