如果当前为空，则插入上一个日期的信息 Oracle答案

【问题标题】：Inserting the previous date's info if the current is empty Oracle如果当前为空，则插入上一个日期的信息 Oracle
【发布时间】：2017-12-24 15:03:08
【问题描述】：

我有一张包含以下信息的表格

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        |

金额是某部门指定货币的账户在该日期当天结束时的金额。更重要的是，Currency 和 Dept_ID（相同的 Account_no 但不同的货币和/或部门 ID）可能会导致同一帐户的变化，我的意思是 PK 是组合Account_no、Dept_ID 和货币。

我正在尝试将该表附加到具有某个指定范围的每天的日期表：

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 Dates 表中的每个日期，我都会从 Table1 获得信息，如果缺少信息，则应选择前几天的信息。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充到缺失的字段中

SELECT * FROM DATES A LEFT JOIN TABLE1 B ON A.DATE_1 = B.DATE_2;

我明白了

|   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 |            |         |         |        |                   |
...

感谢您提供有关如何进行的建议

【问题讨论】：

标签： sql oracle join oracle10g

【解决方案1】：

您可以使用 lag() 和 ignore nulls 选项来执行此操作。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;

【讨论】：

您好，感谢您的宝贵时间，我在使用ignore nulls 时遇到问题，上面写着Found: "ignore" expecting: ) 我试图找到丢失的括号，但没有找到，我做错了什么吗？
@Hatik 。 . . Oracle 绝对支持IGNORE NULLS：docs.oracle.com/cd/E11882_01/server.112/e41084/…。
是的，11支持，我没注意到10.2不支持，所以用last_value代替，应该指定我的版本

【解决方案2】：

您也可以使用partition outer join 加上一个 case 语句来执行此操作，该语句决定是放入当前金额还是如果没有当前金额，则输入以前的可用金额，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

注意如果您使用的是 11.2 之前的 Oracle 版本，则 lag 不会知道忽略空值。您可以改用以下方法来模拟相同的效果：

nvl(last_value(t1.amount) IGNORE NULLS OVER (PARTITION BY t1.account_no, t1.dept_id, t1.currency ORDER BY d.date_1), 0)

【讨论】：

您好！谢谢你的帮助，我想你误解了我一点，如果你的代码中的nvl 不在 table1 中，它将返回 0，但是我想要做的是返回前一天的信息如果当前为空，例如在 05/01/2017 它将返回 amount = 50 为 account_no =1 和 dept_id = 1 因为没有进行任何操作它会与最后一天操作完成时相同
好的，在这种情况下，你需要一个 lag 或 last_value 而不是 nvl。
是的，我也这么认为，这就是我目前正在研究的，感谢@Gordon Linoff
好的，我已经添加了一个应该可以按您预期工作的答案 - 抱歉错过了您的实际要求！
确实，我也得出了同样的结论，Oracle 10g 中没有 IGNORE NULLS 用于延迟，不幸的是我的 db 是 10.2。

【解决方案3】：

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;

【讨论】：