賬戶負6個月或更無論交易

Question

我有幾個帳戶

Select * from Trans; 

AccountID PostDate Description Amount 
1   07/01/2016 deposit  10.00 
1   07/09/2016 withdrawal -15.00 

第二帳戶：

AccountID PostDate Description Amount 
2   07/01/2016 deposit  10.00 
2   07/13/2016 withdrawal -20.00 
2   01/05/2017 deposit  8.00 

第三個帳戶：

AccountID PostDate Description Amount 
3   07/05/2016 deposit  10.00 
3   07/19/2016 deposit  20.00 
3   08/28/2016 withdrawal -45.00 

四帳號：

AccountID PostDate Description Amount 
4   01/05/2016 deposit  10.00 
4   01/19/2016 withdrawal -20.00 
4   09/28/2016 deposit  40.00 
4   10/01/2016 withdrawal -50 

我在尋找所有連續6個月以上爲負數的賬戶，無論存款是否存入，賬戶保持負值。如果存款使運行的餘額爲正，那麼顯然我需要排除該帳戶。

我需要一個通用的查詢......因爲我有超過上述兩個帳戶在trans表中。

查詢應該選擇AccountID 1，因爲它在180天以上爲負數。它應該從2016年7月13日起收取AccountID 2，因爲它是負值。 7月份爲負值-10，雖然有存款，但2017年1月仍保持負值-2。由於餘額爲負值，但在2016年8月28日爲負值，這意味着它不應收回AccountID 3只有148天爲負值。我也不想拿起賬戶'4'。雖然它連續6個月或更長時間爲負值，目前也是負值....但我想要在01/23/2017 - 07/23/2016之間的所有日子裏獲得運行餘額爲負值的帳戶。

謝謝

Answer 1

甲骨文設置：

CREATE TABLE trans (AccountID, PostDate, Description, Amount) AS 
SELECT 1, DATE '2016-07-01', 'deposit',  10.00 FROM DUAL UNION ALL 
SELECT 1, DATE '2016-07-09', 'withdrawal', -15.00 FROM DUAL UNION ALL 
SELECT 2, DATE '2016-07-01', 'deposit',  10.00 FROM DUAL UNION ALL 
SELECT 2, DATE '2016-07-13', 'withdrawal', -20.00 FROM DUAL UNION ALL 
SELECT 2, DATE '2017-01-05', 'deposit',  8.00 FROM DUAL UNION ALL 
SELECT 3, DATE '2016-07-05', 'deposit',  10.00 FROM DUAL UNION ALL 
SELECT 3, DATE '2016-07-19', 'deposit',  20.00 FROM DUAL UNION ALL 
SELECT 3, DATE '2016-08-28', 'withdrawal', -45.00 FROM DUAL UNION ALL 
SELECT 4, DATE '2016-01-05', 'deposit',  10.00 FROM DUAL UNION ALL 
SELECT 4, DATE '2016-01-19', 'withdrawal', -20.00 FROM DUAL UNION ALL 
SELECT 4, DATE '2016-09-28', 'deposit',  40.00 FROM DUAL UNION ALL 
SELECT 4, DATE '2016-10-01', 'withdrawal', -50.00 FROM DUAL; 

查詢：

SELECT accountid 
FROM (
    SELECT t.*, 
     SUM(amount) OVER (PARTITION BY AccountID ORDER BY postdate) 
      AS balance 
    FROM trans t 
) 
GROUP BY accountid 
HAVING MAX(balance) KEEP (DENSE_RANK LAST ORDER BY postdate) < 0 
AND ( MAX(postdate) <= TRUNC(SYSDATE) - 180 
     OR MAX(CASE WHEN postdate >= TRUNC(SYSDATE) - 180 
        THEN balance - amount END) < 0 
     ); 

輸出：

ACCOUNTID 
---------- 
     1 
     2 

Answer 2

首先，你需要計算每一天的餘額。爲此，請使用累積和：

select t.*, 
     sum(amount) over (partition by accountid order by postdate) as balance 
from trans t; 

然後，您希望180天的餘額爲正數或負數。這很棘手。這裏是一個蠻力的方法：

with tb as (
     select t.*, 
      sum(amount) over (partition by accountid order by postdate) as balance 
     from trans t 
    ) 
select distinct accountid 
from tb 
where tb.balance < 0 and 
     not exists (select 1 
        from tb tb2 
        where tb2.accountid = tb.accountid and 
         tb2.balance > 0 and 
         tb2.postdate >= tb.postdate and tb2.postdate < tb.postdate + 180 
       ); 

Answer 3

這是一種解決這個MATCH_RECOGNIZE的方法。它假定PostDate可能沒有任何給定帳戶的重複。 （如果給定日期有多筆交易，交易時間將控制如何處理。）

關鍵是PATTERN子句。我們尋找一系列交易（包括一個賬戶的所有行，因爲我們使用^和$來定位），其中前0行到（未知）行數不受約束（a行），然後我們查找一行（ b），在trunc(SYSDATE)之前爲負值餘額且日期爲180天或更多，其餘所有行必須爲負餘額。

with 
    trans (AccountID, PostDate, Description, Amount) as ( 
     select 1, to_date('07/01/2016', 'mm/dd/yy'), 'deposit' , 10.00 from dual union all 
     select 1, to_date('07/09/2016', 'mm/dd/yy'), 'withdrawal', -15.00 from dual union all 
     select 2, to_date('07/01/2016', 'mm/dd/yy'), 'deposit' , 10.00 from dual union all 
     select 2, to_date('07/13/2016', 'mm/dd/yy'), 'withdrawal', -20.00 from dual union all 
     select 2, to_date('01/05/2017', 'mm/dd/yy'), 'deposit' , 8.00 from dual union all 
     select 3, to_date('07/05/2016', 'mm/dd/yy'), 'deposit' , 10.00 from dual union all 
     select 3, to_date('07/19/2016', 'mm/dd/yy'), 'deposit' , 20.00 from dual union all 
     select 3, to_date('08/28/2016', 'mm/dd/yy'), 'withdrawal', -45.00 from dual union all 
     select 4, to_date('01/05/2016', 'mm/dd/yy'), 'deposit' , 10.00 from dual union all 
     select 4, to_date('01/19/2016', 'mm/dd/yy'), 'withdrawal', -20.00 from dual union all 
     select 4, to_date('09/28/2016', 'mm/dd/yy'), 'deposit' , 40.00 from dual union all 
     select 4, to_date('10/01/2016', 'mm/dd/yy'), 'withdrawal', -50 from dual 
    ) 
-- End of test data; SQL query begins BELOW THIS LINE 
select AccountID 
from trans 
match_recognize(
    partition by AccountID 
    order  by PostDate 
    pattern (^ a* b c* $) 
    define b as b.PostDate <= trunc(sysdate) - 180 and sum(Amount) < 0, 
      c as          sum(Amount) < 0 
) 
; 

ACCOUNTID 
---------- 
     1 
     2 

2 rows selected. 

Answer 4

你也可以試試這個片段。這也足以滿足你的要求。希望這可以幫助。

SELECT b.acc 
FROM 
    (SELECT row_number() over(partition BY a.acc order by dt DESC) nr, 
    SUM(a.amt) over(partition BY a.acc order by 1 DESC) sm, 
    row_number() over(partition BY a.acc,tp order by a.dt DESC) dt1, 
    a.* 
    FROM 
    (SELECT 1 acc, '2016-07-01' dt, 'deposit' tp, 10.00 amt FROM dual 
    UNION ALL 
    SELECT 1, '2016-07-09' dt, 'withdrawal' tp, -15.00 amt FROM dual 
    UNION ALL 
    SELECT 2, '2016-07-01' dt, 'deposit' tp, 10.00 amt FROM dual 
    UNION ALL 
    SELECT 2, '2016-07-13' dt, 'withdrawal' tp, -20.00 amt FROM dual 
    UNION ALL 
    SELECT 2, '2017-01-05' dt, 'deposit' tp, 8.00 amt FROM dual 
    UNION ALL 
    SELECT 3, '2016-07-05' dt, 'deposit' tp, 10.00 amt FROM dual 
    UNION ALL 
    SELECT 3, '2016-07-19' dt, 'deposit' tp, 20.00 amt FROM dual 
    UNION ALL 
    SELECT 3, '2016-08-28' dt, 'withdrawal' tp, -45.00 amt FROM dual 
    UNION ALL 
    SELECT 4, '2016-01-05' dt, 'deposit' tp, 10.00 amt FROM dual 
    UNION ALL 
    SELECT 4, '2016-01-19' dt, 'withdrawal' tp, -20.00 amt FROM dual 
    UNION ALL 
    SELECT 4, '2016-09-28' dt, 'deposit' tp, 40.00 amt FROM dual 
    UNION ALL 
    SELECT 4, '2016-10-01' dt, 'withdrawal' tp, -50.00 amt FROM dual 
    )a 
)b 
WHERE (b.nr          = 1 
OR b.dt1           =1) 
AND b.sm           < 0 
AND b.tp           <> 'deposit' 
AND (TRUNC(sysdate) - to_date(b.dt,'yyyy-mm-dd')) > 180;