我有一张表格，其中包含以下信息
Table1是一个子查询，为了简化，我将只使用结果表：
| Account_No | Dept_ID | Currency| Amount | Date_2(dd/mm/yyyy)|
+------------+---------+---------+--------+-------------------+
| 1 | 1 | USD | 50 | 03/01/2017 |
| 1 | 2 | EUR | 25 | 01/01/2017 |
| 1 | 3 | USD | 51 | 01/01/2017 |
| 1 | 1 | GBP | 45 | 01/01/2017 |
| 1 | 2 | USD | 65 | 02/01/2017 |
金额是某个部门在该日期指定货币的帐户当天结束时的金额。更重要的是，货币和 Dept_ID （相同的帐号，但不同的货币和/或部门ID）可能存在同一帐户的变化我的意思是PK是组合Account_no，Dept_ID和Currency。
我正在尝试将该表附加到日期表中，该日期表具有某些指定范围的日常值：
日期表：
| Date_1 |
+------------+
| 01/01/2017 |
| 02/01/2017 |
| 03/01/2017 |
| 04/01/2017 |
| 05/01/2017 |
...
预期结果是：
| Date_1 | Account_No | Dept_ID | Currency| Amount | Date_2(dd/mm/yyyy)|
+------------+------------+---------+---------+--------+-------------------+
| 01/01/2017 | 1 | 1 | USD | 0 | |
| 01/01/2017 | 1 | 2 | USD | 0 | |
| 01/01/2017 | 1 | 2 | EUR | 25 | 01/01/2017 |
| 01/01/2017 | 1 | 3 | USD | 51 | 01/01/2017 |
| 01/01/2017 | 1 | 1 | GBP | 45 | 01/01/2017 |
| 02/01/2017 | 1 | 2 | USD | 65 | 02/01/2017 |
| 02/01/2017 | 1 | 2 | EUR | 25 | |
| 02/01/2017 | 1 | 3 | USD | 51 | |
| 02/01/2017 | 1 | 1 | GBP | 45 | |
| 02/01/2017 | 1 | 1 | USD | 0 | |
| 03/01/2017 | 1 | 1 | USD | 50 | 03/01/2017 |
| 03/01/2017 | 1 | 1 | GBP | 45 | |
| 03/01/2017 | 1 | 3 | USD | 51 | |
| 03/01/2017 | 1 | 2 | EUR | 25 | |
| 03/01/2017 | 1 | 2 | USD | 65 | |
因此，对于日期表中的每个日期，我都会获得来自 Table1 的信息，如果信息丢失，则应选择前几天＆＃39;信息。我已经完成了左加入的查询，但不知道如何将前一天的数据填入缺失的字段
SELECT * FROM DATES A LEFT JOIN TABLE1 B ON A.DATE_1 = B.DATE_2;
我得到了
| Date_1 | Account_No | Dept_ID | Currency| Amount | Date_2(dd/mm/yyyy)|
+------------+------------+---------+---------+--------+-------------------+
| 01/01/2017 | 1 | 2 | EUR | 25 | 01/01/2017 |
| 01/01/2017 | 1 | 3 | USD | 51 | 01/01/2017 |
| 01/01/2017 | 1 | 1 | GBP | 45 | 01/01/2017 |
| 02/01/2017 | 1 | 2 | USD | 65 | 02/01/2017 |
| 03/01/2017 | 1 | 1 | USD | 50 | 03/01/2017 |
| 04/01/2017 | | | | | |
...
有关如何进行的建议表示赞赏
答案 0 :(得分：1)
您可以使用lag()选项ignore nulls来执行此操作。我想这就是你想要的：
select d.date_1, a.account_no,
coalesce(dept_id,
lag(dept_id ignore nulls) over (partition by t1.account_no order by d.date_1)
) as dept_id,
coalesce(currency,
lag(currency ignore nulls) over (partition by t1.account_no order by d.date_1)
) as currency,
coalesce(amount,
lag(amount ignore nulls) over (partition by t1.account_no order by d.date_1)
) as amount
from dates d CROSS JOIN
(select distinct account_no from table1) a left join
table1 t1
on d.DATE_1 = t1.DATE_2 and a.account_no = t1.account_no;
答案 1 :(得分：1)
您也可以使用partition outer join加上一个案例陈述来决定是否在没有当前金额的情况下存入当前金额或之前的可用金额，如下所示：
WITH table1 AS (SELECT 1 account_no, 1 dept_id, 'USD' currency, 50 amount, to_date('03/01/2017', 'dd/mm/yyyy') date_2 FROM dual UNION ALL
SELECT 1 account_no, 2 dept_id, 'EUR' currency, 25 amount, to_date('01/01/2017', 'dd/mm/yyyy') date_2 FROM dual UNION ALL
SELECT 1 account_no, 3 dept_id, 'USD' currency, 51 amount, to_date('01/01/2017', 'dd/mm/yyyy') date_2 FROM dual UNION ALL
SELECT 1 account_no, 1 dept_id, 'GBP' currency, 45 amount, to_date('01/01/2017', 'dd/mm/yyyy') date_2 FROM dual UNION ALL
SELECT 1 account_no, 2 dept_id, 'USD' currency, 65 amount, to_date('02/01/2017', 'dd/mm/yyyy') date_2 FROM dual),
dates AS (SELECT to_date('01/01/2017', 'dd/mm/yyyy') date_1 FROM dual UNION ALL
SELECT to_date('02/01/2017', 'dd/mm/yyyy') date_1 FROM dual UNION ALL
SELECT to_date('03/01/2017', 'dd/mm/yyyy') date_1 FROM dual UNION ALL
SELECT to_date('04/01/2017', 'dd/mm/yyyy') date_1 FROM dual UNION ALL
SELECT to_date('05/01/2017', 'dd/mm/yyyy') date_1 FROM dual)
SELECT d.date_1,
t1.account_no,
t1.dept_id,
t1.currency,
CASE WHEN t1.amount is NULL THEN
LAG(t1.amount, 1, 0) IGNORE NULLS OVER (PARTITION BY t1.account_no, t1.dept_id, t1.currency ORDER BY d.date_1)
ELSE t1.amount
END amount,
t1.date_2
FROM dates d
LEFT OUTER JOIN table1 t1 PARTITION BY (t1.account_no, t1.dept_id, t1.currency)
ON d.date_1 = t1.date_2
ORDER BY d.date_1,
t1.account_no,
t1.dept_id,
t1.currency;
DATE_1 ACCOUNT_NO DEPT_ID CURRENCY AMOUNT DATE_2
----------- ---------- ---------- -------- ---------- -----------
01/01/2017 1 1 GBP 45 01/01/2017
01/01/2017 1 1 USD 0
01/01/2017 1 2 EUR 25 01/01/2017
01/01/2017 1 2 USD 0
01/01/2017 1 3 USD 51 01/01/2017
02/01/2017 1 1 GBP 45
02/01/2017 1 1 USD 0
02/01/2017 1 2 EUR 25
02/01/2017 1 2 USD 65 02/01/2017
02/01/2017 1 3 USD 51
03/01/2017 1 1 GBP 45
03/01/2017 1 1 USD 50 03/01/2017
03/01/2017 1 2 EUR 25
03/01/2017 1 2 USD 65
03/01/2017 1 3 USD 51
04/01/2017 1 1 GBP 45
04/01/2017 1 1 USD 50
04/01/2017 1 2 EUR 25
04/01/2017 1 2 USD 65
04/01/2017 1 3 USD 51
05/01/2017 1 1 GBP 45
05/01/2017 1 1 USD 50
05/01/2017 1 2 EUR 25
05/01/2017 1 2 USD 65
05/01/2017 1 3 USD 51
N.B。如果您使用的是11.2之前的Oracle版本，则lag不会知道忽略空值。您可以使用以下代码来模拟相同的效果：
nvl(last_value(t1.amount) IGNORE NULLS OVER (PARTITION BY t1.account_no, t1.dept_id, t1.currency ORDER BY d.date_1), 0)
答案 2 :(得分：0)
WITH table1 AS
(
SELECT 1 account_no,
1 dept_id,
'USD' currency,
50 amount,
to_date('03/01/2017', 'dd/mm/yyyy') date_2
FROM dual
UNION ALL
SELECT 1 account_no,
2 dept_id,
'EUR' currency,
25 amount,
to_date('01/01/2017', 'dd/mm/yyyy') date_2
FROM dual
UNION ALL
SELECT 1 account_no,
3 dept_id,
'USD' currency,
51 amount,
to_date('01/01/2017', 'dd/mm/yyyy') date_2
FROM dual
UNION ALL
SELECT 1 account_no,
1 dept_id,
'GBP' currency,
45 amount,
to_date('01/01/2017', 'dd/mm/yyyy') date_2
FROM dual
UNION ALL
SELECT 1 account_no,
2 dept_id,
'USD' currency,
65 amount,
to_date('02/01/2017', 'dd/mm/yyyy') date_2
FROM dual
)
,
dates AS
(
SELECT to_date('01/01/2017', 'dd/mm/yyyy') date_1 FROM dual
UNION ALL
SELECT to_date('02/01/2017', 'dd/mm/yyyy') date_1 FROM dual
UNION ALL
SELECT to_date('03/01/2017', 'dd/mm/yyyy') date_1 FROM dual
UNION ALL
SELECT to_date('04/01/2017', 'dd/mm/yyyy') date_1 FROM dual
UNION ALL
SELECT to_date('05/01/2017', 'dd/mm/yyyy') date_1 FROM dual
)
SELECT d.date_1,
t1.account_no,
t1.dept_id,
t1.currency,
t1.amount,
CASE (t1.date_2)
WHEN d.date_1
THEN t1.date_2
ELSE NULL
END
FROM table1 t1,
dates d
ORDER BY d.date_1,
t1.account_no,
t1.dept_id,
t1.currency;