给出下表
PAYMENT_Date TRANSACTION_TYPE PAYMENT_AMT
1/1/2012 P 184366
1/1/2012 R -5841
1/2/2012 P 941
1/3/2012 P 901
1/3/2012 R 5841
以及以下查询：
select payment_date, transaction_type, payment_amt,
SUM(payment_amt) OVER(ORDER BY payment_date, transaction_type
RANGE BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) AS RUNNING_BALANCE
from TABLE;
我得到了这些结果：
PAYMENT_Date TRANSACTION_TYPE PAYMENT_AMT RUNNING_BALANCE
1/1/2012 P 184366 0
1/1/2012 R -5841 -184366
1/2/2012 P 941 -178525
1/3/2012 P 901 -179466
1/3/2012 R 5841 -180367
预期：
PAYMENT_Date TRANSACTION_TYPE PAYMENT_AMT RUNNING_BALANCE
1/1/2012 P 184366 0
1/1/2012 R -5841 184366
1/2/2012 P 941 178525
1/3/2012 P 901 179466
1/3/2012 R 5841 180367
为什么RUNNING_BALANCE以负数形式返回？ 除了明显的abs（）之外，我怎么能做到呢？
答案 0 :(得分：6)
首先，您发布的数据和查询似乎不会生成您看到的输出。所以某处有某种复制和粘贴错误
SQL> with t as (
2 select date '2012-01-01' payment_date, 'P' transaction_type, 184366 payment_amt from dual union all
3 select date '2012-01-01', 'R', -5841 from dual union all
4 select date '2012-01-02', 'P', 941 from dual union all
5 select date '2012-01-03', 'P', 901 from dual union all
6 select date '2012-01-03', 'R', 5841 from dual
7 )
8 select payment_date, transaction_type, payment_amt,
9 SUM(payment_amt) OVER(ORDER BY payment_date, transaction_type
10 RANGE BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) AS RUNNING_BALANCE
11 from T;
PAYMENT_D T PAYMENT_AMT RUNNING_BALANCE
--------- - ----------- ---------------
01-JAN-12 P 184366 186208
01-JAN-12 R -5841 1842
02-JAN-12 P 941 7683
03-JAN-12 P 901 6742
03-JAN-12 R 5841 5841
通常，只需省略RANGE BETWEEN子句即可完成运行平衡。
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select date '2012-01-01' payment_date, 'P' transaction_type, 184366 payment_amt from dual union all
3 select date '2012-01-01', 'R', -5841 from dual union all
4 select date '2012-01-02', 'P', 941 from dual union all
5 select date '2012-01-03', 'P', 901 from dual union all
6 select date '2012-01-03', 'R', 5841 from dual
7 )
8 select payment_date, transaction_type, payment_amt,
9 SUM(payment_amt) OVER(ORDER BY payment_date, transaction_type) AS RUNNING_BALANCE
10* from T
SQL> /
PAYMENT_D T PAYMENT_AMT RUNNING_BALANCE
--------- - ----------- ---------------
01-JAN-12 P 184366 184366
01-JAN-12 R -5841 178525
02-JAN-12 P 941 179466
03-JAN-12 P 901 180367
03-JAN-12 R 5841 186208
但是，在您的情况下，听起来您希望运行余额排除当前行的付款。这有点奇怪，你可以通过添加额外的LAG
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select date '2012-01-01' payment_date, 'P' transaction_type, 184366 payment_amt from dual union all
3 select date '2012-01-01', 'R', -5841 from dual union all
4 select date '2012-01-02', 'P', 941 from dual union all
5 select date '2012-01-03', 'P', 901 from dual union all
6 select date '2012-01-03', 'R', 5841 from dual
7 )
8 select payment_date,
9 transaction_type,
10 payment_amt,
11 NVL( LAG(running_balance) OVER(ORDER BY payment_date,
12 transaction_type), 0) new_running_balance
13 from (select payment_date,
14 transaction_type,
15 payment_amt,
16 SUM(payment_amt) OVER(ORDER BY payment_date,
17 transaction_type) AS RUNNING_BALANCE
18* from t)
SQL> /
PAYMENT_D T PAYMENT_AMT NEW_RUNNING_BALANCE
--------- - ----------- -------------------
01-JAN-12 P 184366 0
01-JAN-12 R -5841 184366
02-JAN-12 P 941 178525
03-JAN-12 P 901 179466
03-JAN-12 R 5841 180367
答案 1 :(得分：5)
我认为你需要改变：
RANGE BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING`
为：
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
在SQLfiddle中进行测试：
SELECT payment_date, transaction_type, payment_amt,
COALESCE( SUM(payment_amt)
OVER( ORDER BY payment_date, transaction_type
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING)
, 0) AS RUNNING_BALANCE
FROM T;