Question

我有一个表，用于存储oracle中客户回复的状态。我必须计算客户的最后连续下降。
例如：
身份证明状态
-----------------------------
1.拒绝了 2.接受的 3.拒绝了 4.拒绝了

这将有计数= 2.
因为最后两个被拒绝了。

Answer 1

注意：原始海报在“评论”中澄清，她需要一个不同的要求 - 在一个单独的答案中解决它。保持这一点，因为它显示了OP的更复杂问题的一种可能的解决方案。

假设您想要计算最近连续的“下降”（即使后面跟着“已接受” - 并且您希望允许多个客户 - 这是一种可能的解决方案。

输入表格（我称之为“t”）：

SQL> select * from t;

CUSTOMER_ID DECISION_ID STATUS
----------- ----------- --------------------
         10           1 Accepted
         10           2 Declined
         10           3 Declined
         10           4 Accepted
         10           5 Accepted
         10           6 Declined
         10           7 Declined
         30           1 Declined
         30           2 Accepted
         30           3 Declined
         30           4 Accepted
         30           5 Declined
         30           6 Declined
         30           7 Declined
         30           8 Accepted
         30           9 Accepted

<强>查询：

with t1 as 
  (
    select customer_id, 
           decision_id - row_number() over 
                             (partition by customer_id order by decision_id) as idx 
    from   t 
    where  status = 'Declined'
  ),
  t2 as (select customer_id, max(idx) as max_idx from t1 group by customer_id)
select t1.customer_id, count(1) as ct
from t1 join t2 on t1.customer_id = t2.customer_id 
where t1.idx = t2.max_idx
group by t1.customer_id
order by t1.customer_id
/

查询输出：

CUSTOMER_ID         CT
----------- ----------
         10          2
         30          3

Answer 2

@Gurmeet - 然后问题就容易多了。这是解决它的一种方法。如果您需要customer_id排序的结果，请在最后添加order by customer_id。如果客户从未进行过具有“已接受”状态的交易，则需要在CTE t2中定义d_A中的nvl。

输入：与我的其他答案相同。

查询 :(已修改以满足OP附加要求）：

    with t0 as (select customer_id, status, row_number() over 
                        (partition by customer_id order by decision_id) rn from t),
         t1 as (select customer_id, max(rn) as d_all from t0 group by customer_id),
         t2 as (select customer_id, nvl(max(rn), 0) as d_A from t0
                where status = 'Accept' group by customer_id)
    select customer_id, d_all - d_A as ct from t1 natural join t2

    /

<强>结果：

CUSTOMER_ID         CT
----------- ----------
         30          0
         10          2

2 rows selected.

oracle：计算连续行中状态的出现次数

2 个答案: