Oracle SQL 层次结构求和答案

【问题标题】：Oracle SQL Hierarchy SummationOracle SQL 层次结构求和
【发布时间】：2020-08-18 08:46:54
【问题描述】：

我有一个包含以下记录的表 TRANS：

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 的初始余额为 0（2020 年 8 月 1 日）。对于随后的日子，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。是否有可能在 1 个 SQL 语句中获得输出？

谢谢。

【问题讨论】：

标签： sql oracle sum window-functions recursive-query

【解决方案1】：

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

【讨论】：

【解决方案2】：

你也可以这样做（我认为它会快一点）：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 在您的数据中返回min(trans_dt) 和max(trans_dt) 之间的所有天数。

dt_for_trans_id 返回数据中每个 trans_id 的所有这些天数。

qty_change 查找每天的差异（在您的示例中为 TOTAL）和所有天的累积总和（在您的示例中为 END_BAL）。

主选择从前一天获取END_BAL，并将其命名为BEG_BAL，它还对最终输出进行了一些格式化。

【讨论】：

【解决方案3】：

首先，您需要生成日期，然后您需要通过 TRANS_DT 聚合您的值，然后将聚合的数据左连接到日期。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;

带有样本数据的完整示例：

alter session set nls_date_format='dd-mon-yyyy';
with trans(TRANS_ID,TRANS_DT,QTY) as (
select 1,to_date('01-Aug-2020'),  5 from dual union all
select 1,to_date('01-Aug-2020'),  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_agg as ( -- aggregating QTY in TRANS
   select TRANS_ID,TRANS_DT,sum(QTY) as QTY
   from trans
   group by TRANS_ID,TRANS_DT
)
select 
   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;

【讨论】：

【解决方案4】：

您可以使用递归查询来生成整个日期范围，cross join 它带有不同的tran_id 列表，然后将带有left join 的表带入。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,
    coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 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

【讨论】：