Question

现在，我在MS Access中有一个表X，其中包含以下字段：

X (Table)
- Dates   (Date/Time)
- Number1 (Double)
- Number2 (Long Integer)

现在假设它有以下三个记录：

2000-01-01 00:00:00,0.3,4
2000-01-01 00:02:00,0.5,5
2000-01-01 00:05:00,0.8,4

我想编写一个可以通过OLE DB调用的视图或其他内容来返回以下内容：

2000-01-01 00:00:00,0.3,4
2000-01-01 00:01:00,0.5,5
2000-01-01 00:02:00,0.5,5
2000-01-01 00:03:00,0.8,4
2000-01-01 00:04:00,0.8,4
2000-01-01 00:05:00,0.8,4

我希望用下一个可用数据“填写”缺少的会议记录。有可能吗？

Answer 1

对于纯SQL解决方案，我们可以使用[Numbers]表和一个Long Integer列，其中至少值为0到1439，例如，

Integer
-------
      0
      1
      2
      3
...
   1439

我们可以在Access中创建一个名为 [X_with_minutes] 的已保存查询，该查询检索[X]表中的数据，并添加两列显示：（1）只是日期，（2））午夜过后的分钟数......

SELECT X.*, 
    DateSerial(Year([Dates]), Month([Dates]), Day([Dates])) AS JustDate, 
    DateDiff("n", DateSerial(Year([Dates]), Month([Dates]), Day([Dates])), [Dates]) AS DayMinutes
FROM X;

...返回：

Dates                Number1  Number2  JustDate    DayMinutes
-------------------  -------  -------  ----------  ----------
2000-01-01 00:00:00      0.3        4  2000-01-01           0
2000-01-01 00:02:00      0.5        5  2000-01-01           2
2000-01-01 00:05:00      0.8        4  2000-01-01           5

我们可以在Access中创建另一个名为 [X_minutes_all] 的已保存查询，以便在[X]中生成每一天的所有分钟......

SELECT DISTINCT X_with_minutes.JustDate, Numbers.Integer AS DayMinutes
FROM X_with_minutes, Numbers
WHERE Numbers.Integer BETWEEN 0 AND 1439;

...返回：

JustDate    DayMinutes
----------  ----------
2000-01-01           0
2000-01-01           1
2000-01-01           2
2000-01-01           3
...
2000-01-01        1439

现在我们可以创建一个返回缺失分钟的查询，并将该查询保存为 [X_missing_minutes]

SELECT * FROM  X_Minutes_all 
WHERE NOT EXISTS 
    (
        SELECT * FROM X_with_minutes 
        WHERE X_with_minutes.JustDate=X_minutes_all.JustDate 
            AND X_with_minutes.DayMinutes=X_minutes_all.DayMinutes
    );

...返回：

JustDate    DayMinutes
----------  ----------
2000-01-01           1
2000-01-01           3
2000-01-01           4
2000-01-01           6
2000-01-01           7
2000-01-01           8
...
2000-01-01        1439

要找到缺失分钟的“下一个”分钟值，我们需要找到超过当前值的最小（MIN）值。我们可以按如下方式创建该查询，并将其另存为 [X_next_minutes] ...

SELECT X_missing_minutes.JustDate, 
    X_missing_minutes.DayMinutes,
    (
        SELECT MIN(DayMinutes) 
        FROM X_with_minutes 
        WHERE X_with_minutes.JustDate=X_missing_minutes.JustDate 
            AND X_with_minutes.DayMinutes>X_missing_minutes.DayMinutes
    ) AS NextMinute
FROM X_missing_minutes;

...返回...

JustDate    DayMinutes  NextMinute
----------  ----------  ----------
2000-01-01           1           2
2000-01-01           3           5
2000-01-01           4           5
2000-01-01           6      <NULL>
2000-01-01           7      <NULL>
2000-01-01           8      <NULL>
...
2000-01-01        1439      <NULL>

我们可以使用它来返回缺少的分钟的[X]值...

SELECT DateAdd("n", X_next_minutes.DayMinutes, X_next_minutes.JustDate) AS Dates, 
    X_with_minutes.Number1, X_with_minutes.Number2 
FROM X_next_minutes INNER JOIN X_with_minutes 
    ON X_next_minutes.JustDate=X_with_minutes.JustDate 
        AND X_next_minutes.NextMinute=X_with_minutes.DayMinutes;

...返回...

Dates                Number1  Number2
-------------------  -------  -------
2000-01-01 00:01:00      0.5        5
2000-01-01 00:03:00      0.8        4
2000-01-01 00:04:00      0.8        4

...我们可以使用原始[X]来加强UNION ALL以产生最终结果......

SELECT DateAdd("n", X_next_minutes.DayMinutes, X_next_minutes.JustDate) AS Dates, 
    X_with_minutes.Number1, X_with_minutes.Number2 
FROM X_next_minutes INNER JOIN X_with_minutes 
    ON X_next_minutes.JustDate=X_with_minutes.JustDate 
        AND X_next_minutes.NextMinute=X_with_minutes.DayMinutes
UNION ALL
SELECT * FROM X
ORDER BY 1;

...返回...

Dates                Number1  Number2
-------------------  -------  -------
2000-01-01 00:00:00      0.3        4
2000-01-01 00:01:00      0.5        5
2000-01-01 00:02:00      0.5        5
2000-01-01 00:03:00      0.8        4
2000-01-01 00:04:00      0.8        4
2000-01-01 00:05:00      0.8        4

零时间保持......可能吗？

1 个答案: