Oracle SQL层次结构汇总

如何解决Oracle SQL层次结构汇总

我有一个TRANS表，其中包含以下记录：

TRANS_ID TRANS_DT     QTY    
1        01-Aug-2020  5
1        01-Aug-2020  1
1        03-Aug-2020  2
2        02-Aug-2020  1

预期输出：

TRANS_ID TRANS_DT     BEGBAL TOTAL END_BAL 
1        01-Aug-2020  0      6     6
1        02-Aug-2020  6      0     6      
1        03-Aug-2020  6      2     8
2        01-Aug-2020  0      0     0
2        02-Aug-2020  0      1     1      
2        03-Aug-2020  1      0     1

每个trans_id的起始余额为0（2020年8月1日）。对于后续的几天，期初余额是前一天的期末余额，依此类推。我可以创建PL / SQL块来创建输出。是否可以在1条SQL语句中获得输出？

谢谢。

解决方法

使用CTE尝试以下脚本-

Demo Here

WITH CTE 
AS
(
    SELECT DISTINCT A.TRANS_ID,B.TRANS_DT
    FROM your_table A
    CROSS JOIN (SELECT DISTINCT TRANS_DT FROM your_table) B

),CTE2
AS
(
    SELECT C.TRANS_ID,C.TRANS_DT,SUM(D.QTY) QTY
    FROM CTE C
    LEFT JOIN your_table D 
        ON C.TRANS_ID = D.TRANS_ID 
        AND C.TRANS_DT = D.TRANS_DT
    GROUP BY C.TRANS_ID,C.TRANS_DT
    ORDER BY C.TRANS_ID,C.TRANS_DT
)

SELECT F.TRANS_ID,F.TRANS_DT,(    
    SELECT COALESCE (SUM(QTY),0) FROM CTE2 E 
    WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT < F.TRANS_DT
) BEGBAL,0) FROM CTE2 E 
    WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT = F.TRANS_DT
) TOTAL,0) FROM CTE2 E 
    WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT <= F.TRANS_DT
) END_BAL 
FROM CTE2 F

您也可以这样做（我想它会快一点）：Demo

with 
dt_between as (
  select mindt + level - 1 as trans_dt
  from (select min(trans_dt) as mindt,max(trans_dt) as maxdt from t)
  connect by level <= maxdt - mindt + 1
),dt_for_trans_id as (
  select *
  from dt_between,(select distinct trans_id from t)
),qty_change as (
  select distinct trans_id,trans_dt,sum(qty) over (partition by trans_id,trans_dt) as total,sum(qty) over (partition by trans_id order by trans_dt) as end_bal
  from t
  right outer join dt_for_trans_id using (trans_id,trans_dt)
)
select 
  trans_id,to_char(trans_dt,'DD-Mon-YYYY') as trans_dt,nvl(lag(end_bal) over (partition by trans_id order by trans_dt),0) as beg_bal,nvl(total,0) as total,nvl(end_bal,0) as end_bal
from qty_change q
order by trans_id,trans_dt

dt_between返回数据中min(trans_dt)和max(trans_dt)之间的所有日期。

dt_for_trans_id将为您数据中的每个trans_id返回所有这些天。

qty_change会发现每天的差异（在您的示例中为TOTAL）和所有天的累计金额（在示例中为END_BAL）。

主选择从前一天取END_BAL并命名为BEG_BAL，它还会对最终输出进行一些格式化。

首先，您需要生成日期，然后需要通过TRANS_DT汇总您的值，然后使汇总数据与日期保持连接。获得所需总和的最简单方法是使用分析窗函数：

with dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
   select min(trans_dt) from trans
   union all
   select dt+1 from dates
   where dt+1<=(select max(trans_dt) from trans)
),trans_agg as ( -- aggregating QTY in TRANS
   select TRANS_ID,TRANS_DT,sum(QTY) as QTY
   from trans
   group by TRANS_ID,TRANS_DT
)
select -- using left join partition by to get data on daily basis for each trans_id:
   dt,trans_id,nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,nvl(qty,0) as TOTAL,nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL 
from dates
     left join trans_agg tr
          partition by (trans_id)
          on tr.trans_dt=dates.dt;

带有示例数据的完整示例：

alter session set nls_date_format='dd-mon-yyyy';
with trans(TRANS_ID,QTY) as (
select 1,to_date('01-Aug-2020'),5 from dual union all
select 1,1 from dual union all
select 1,to_date('03-Aug-2020'),2 from dual union all
select 2,to_date('02-Aug-2020'),1 from dual
),dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
   select min(trans_dt) from trans
   union all
   select dt+1 from dates
   where dt+1<=(select max(trans_dt) from trans)
),TRANS_DT
)
select 
   dt,0) as END_BAL 
from dates
     left join trans_agg tr
          partition by (trans_id)
          on tr.trans_dt=dates.dt;

您可以使用递归查询来生成整个日期范围，将其与单独的cross join列表一起tran_id，然后将表中带有left join。最后一步是聚合和窗口函数：

with all_dates (trans_dt,max_dt) as (
    select min(trans_dt),max(trans_dt) from trans group by trans_id
    union all
    select trans_dt + interval '1' day,max_dt from all_dates where trans_dt < max_dt
)
select
    i.trans_id,d.trans_dt,coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt),0) - coalesce(sum(t.qty),0) begbal,coalesce(sum(t.qty),0) total,0) endbal
from all_dates d
cross join (select distinct trans_id from trans) i
left join trans t on t.trans_id = i.trans_id and t.trans_dt = d.trans_dt
group by i.trans_id,d.trans_dt
order by i.trans_id,d.trans_dt

Oracle SQL层次结构汇总

如何解决Oracle SQL层次结构汇总

解决方法

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