在Oracle中根据多组7个数字范围计算计数答案

【问题标题】：Calculate count based on multiple groups of 7 number range in Oracle在Oracle中根据多组7个数字范围计算计数
【发布时间】：2020-01-30 23:05:02
【问题描述】：

谁能帮帮我，我卡在这里了。

给定

AMOUNT  CUSTOMER_NUMBER GROUPID SEQF
1000+   5555            51      2       
1000+   5555            52      3
1000+   5555            55      4   
1000+   5555            56      4   
1000+   5555            57      4       
1000+   5555            58      2   
1000+   5555            59      4   
1000+   5555            61      2
2000+   6666            55      2
.
.   
.
.

通过考虑 AMOUNT 和 CUSTOMER_NUMBER 列，这里在 7 个 GROUPID 数字内创建组

例如第 1 组为 5555 CUSTOMER_NUMBER ->（51 到 57 组 ID），第 2 组为 5555 CUSTOMER_NUMBER（58 到 61 组 ID），第 3 组为 6666 CUSTOMER_NUMBER（55 组 ID）

需要：

如果 SEQF >=2 在 51 到 57 的任何 GROUPID 中的第一个组中，则考虑 COUNT(SEQF) = 1

如果 SEQF >=2 在 58 到 61 的任何 GROUPID 中的第 2 组中，则考虑 COUNT(SEQF) = 1

这里很全面

AMOUNT  CUSTOMER_NUMBER COUNT(SEQF)
1000+       5555                2

如果 SEQF >=2 在第 3 组中的任何 GROUPID 中的 55 中，则考虑 COUNT(SEQF) = 1

AMOUNT  CUSTOMER_NUMBER COUNT(SEQF)
2000+       6666                1

期望的输出

AMOUNT  CUSTOMER_NUMBER COUNT(SEQF)
1000+       5555                2
2000+       6666                1

【问题讨论】：

您使用的是 Oracle 还是 MySQL？你标记了两者，也许是偶然的？如果你用 MySQL，你用的是什么版本？
groupid 的组是否已固定，例如第一组总是51到57？还是按groupid 的顺序只装7 个？如果有，是否有差距，如果有差距怎么办——无论差距与否，都取 7？
是的，它是 Oracle。我错误地标记了mysql。它是按 groupid 顺序排列的 7 包。另外，我们必须考虑到要取 7 的差距。

标签： sql oracle count grouping

【解决方案1】：

其他答案都没有解决组之间存在差距的问题（他们只是假设组是连续的）。这将跳过间隙并将groupid 放入 7 的范围内。

Oracle 设置：

create table data(amount, customer_number, groupid, seqf) as (
    select '1000+', 5555, 51, 2 from dual union all
    select '1000+', 5555, 52, 3 from dual union all
    select '1000+', 5555, 55, 4 from dual union all
    select '1000+', 5555, 56, 4 from dual union all
    select '1000+', 5555, 57, 4 from dual union all
    select '1000+', 5555, 60, 2 from dual union all
    select '1000+', 5555, 61, 4 from dual union all
    select '1000+', 5555, 65, 2 from dual union all
    select '1000+', 5555, 69, 2 from dual union all
    select '2000+', 6666, 55, 2 from dual );

查询 1：

这会找到组：

SELECT amount,
       customer_number,
       groupid,
       groupid + 6 AS max_groupid
FROM   (
  SELECT d.*,
         MIN( groupid )
           OVER ( PARTITION BY amount, customer_number )
           AS min_groupid,
         MIN( groupid )
           OVER (
             PARTITION BY amount, customer_number
             ORDER BY groupid
             RANGE BETWEEN 7 FOLLOWING AND UNBOUNDED FOLLOWING
           )
           AS next_groupid
  FROM   data d
)
START WITH groupid = min_groupid
CONNECT BY PRIOR amount          = amount
AND        PRIOR customer_number = customer_number
AND        PRIOR next_groupid    = groupid

哪个输出：

数量 | CUSTOMER_NUMBER |组ID | MAX_GROUPID :----- | --------------: | ------: | ----------: 1000+ | 5555 | 51 | 57 1000+ | 5555 | 60 | 66 1000+ | 5555 | 69 | 75 2000+ | 6666 | 55 | 61

查询 2：

这会计算组：

SELECT amount,
       customer_number,
       COUNT(*)
FROM   (
  SELECT d.*,
         MIN( groupid )
           OVER ( PARTITION BY amount, customer_number )
           AS min_groupid,
         MIN( groupid )
           OVER (
             PARTITION BY amount, customer_number
             ORDER BY groupid
             RANGE BETWEEN 7 FOLLOWING AND UNBOUNDED FOLLOWING
           )
           AS next_groupid
  FROM   data d
)
START WITH groupid = min_groupid
CONNECT BY PRIOR amount          = amount
AND        PRIOR customer_number = customer_number
AND        PRIOR next_groupid    = groupid
GROUP BY amount, customer_number

哪个输出：

数量 | CUSTOMER_NUMBER |数数（*） :----- | --------------: | --------: 1000+ | 5555 | 3 2000+ | 6666 | 1

db小提琴here

【讨论】：

【解决方案2】：

我认为您需要将 7 条记录分组并找到一些详细信息，然后再次按客户和金额分组以获得最终结果。

一种方法是：

Select amount, customer_number, sum(seqf) as seqf 
from
  (Select amount, customer_number, 
          case when sum(case when seqf>= 2 then 1 end) >= 1 then 1 end as seqf 
     from
       (Select t.*, 
               Row_number() over(partition by amount, customer_number order by groupid) as rn
          From your_table t)
        Group by amount, customer_number, trunc((rn-1)/7))
Group by amount, customer_number;

干杯！！

【讨论】：

【解决方案3】：

首先你必须找到 7 个 GROUPID 范围，例如使用这个分层查询：

with ranges(amt, cn, g1, g2, mxg) as (
  select amount, customer_number, min(groupid), min(groupid) + 6, max(groupid)
    from data group by amount, customer_number union all
  select amt, cn, g1 + 7, g2 + 7, mxg from ranges where g1 + 7 <= mxg)
select amt, cn, g1, g2, mxg from ranges

这给了我们每个范围（G1，G2）的下限和上限：

AMT           CN         G1         G2 
----- ---------- ---------- ---------- 
1000+       5555         51         57 
1000+       5555         58         64 
2000+       6666         55         61

现在我们可以将这些范围与您的数据连接起来，有条件地计数并求和：

with ranges(amt, cn, g1, g2, mxg) as (
    select amount, customer_number, min(groupid), min(groupid) + 6, max(groupid)
      from data group by amount, customer_number union all
    select amt, cn, g1 + 7, g2 + 7, mxg from ranges where g1 + 7 <= mxg)
select amt, cn, sum(cs) cnt
  from (
    select amt, cn, g1, g2, case when max(seqf) >= 2 then 1 end cs
      from ranges
      join data on amt = amount and cn = customer_number 
               and groupid between g1 and g2
      group by amt, cn, g1, g2)
  group by amt, cn

^{dbfiddle demo}

我使用了max(seqf)，因为您在任何 GROUPID 中的第一个组中的 IF SEQF >=2 表明了这一点。例如，如果您想要该组中的第一个值，请使用 min ... keep dense rank first。

您还可以为每个 (amount, customer_number) 减去 groupid 和 minimum groupid，除以 7，使用 floor，这将创建组，这样会更快：

select amount, customer_number, sum(seqf)
  from (
    select amount, customer_number, case when max(seqf) >= 2 then 1 end seqf
      from (
        select data.*, 
               floor((groupid - min(groupid) over (partition by amount, customer_number))/7) gid
          from data)
      group by amount, customer_number, gid)
  group by amount, customer_number

【讨论】：