Question

我有一个包含以下结构的表

ID     Account Number     Date
1      1001               10/9/2011 (dd/mm/yyyy)
2      2001               1/9/2011 (dd/mm/yyyy)
3      2001               3/9/2011 (dd/mm/yyyy)
4      1001               12/9/2011 (dd/mm/yyyy)
5      3001               18/9/2011 (dd/mm/yyyy)
6      1001               20/9/2011 (dd/mm/yyyy)

基本上我想要做的是有一个访问查询来计算连续记录的日期差异但是对于相同的帐号预期的结果将是!!

1001      10/9/2011 - 12/9/2011     2 days
1001      12/9/2011 - 20/9/2011     8 days
1001      20/9/2011                 NA

基本上我想要做的是有一个访问查询来计算连续记录的日期差异但是对于相同的帐号，在上面的示例中将是1001.（日期不必显示在结果）

我使用access 2003。

Answer 1

SELECT  T1.ID, 
        T1.AccountNumber, 
        T1.Date, 
        MIN(T2.Date) AS Date2, 
        DATEDIFF("D", T1.Date, MIN(T2.Date)) AS DaysDiff
FROM    YourTable T1
        LEFT JOIN YourTable T2
            ON T1.AccountNumber = T2.Accountnumber
            AND T2.Date > T1.Date
GROUP BY T1.ID, T1.AccountNumber, T1.Date;

或

SELECT  ID,
        AccountNumber,
        Date,
        NextDate,
        DATEDIFF("D", Date, NextDate)
FROM    (   SELECT  ID, 
                    AccountNumber,
                    Date,
                    (   SELECT  MIN(Date) 
                        FROM    YourTable T2
                        WHERE   T2.Accountnumber = T1.AccountNumber
                        AND     T2.Date > T1.Date
                    ) AS NextDate
            FROM    YourTable T1
        ) AS T

Answer 2

如果需要，您可以为帐号添加WHERE语句。您的表名为t4

SELECT 
   t4.ID, 
   t4.AccountNumber, 
   t4.AcDate, 
   (SELECT TOP 1 AcDate 
    FROM t4 b 
    WHERE b.AccountNumber=t4.AccountNumber And b.AcDate>t4.AcDate 
    ORDER BY AcDate DESC, ID) AS NextDate, 
   [NextDate]-[AcDate] AS Diff
FROM t4
ORDER BY t4.AcDate;

Answer 3

试试这个：

select [Account Number], DATEDIFF(DD, min(date), max(date)) as dif
from your_table
group by [Account Number]

Answer 4

GarethD的回答对我很有帮助。

仅供参考：当您需要ORDER BY子句时，请在根目录中的SELECT查询结束时使用它。

SELECT  ConsignorID,
            DateRequired StartDate,
            NextDate,
            DATEDIFF("D", DateRequired, NextDate)
FROM (  SELECT  ConsignorID,
                DateRequired,
                (SELECT MIN(DateRequired) 
                 FROM "TABLENAME" T2
                 WHERE T2.DateRequired > T1.DateRequired
                ) AS NextDate
            FROM "TABLENAME" T1
        ) AS T

ORD BY T.DateRequired ASC

Answer 5

您还可以使用LAG分析功能来获得所需结果，如下所示：

假设下面是您的输入表：

id  account_number  account_date
1     1001          9/10/2011
2     2001          9/1/2011
3     2001          9/3/2011
4     1001          9/12/2011
5     3001          9/18/2011
6     1001          9/20/2011


select id,account_number,account_date,
datediff(day,lag(account_date,1) over (partition by account_number order by account_date asc),account_date)
as day_diffrence
from yourtable;

这是您的输出：

id  account_number  account_date    day_diffrence
1     1001           9/10/2011    NULL
4     1001           9/12/2011    2
6     1001           9/20/2011    8
2     2001           9/1/2011     NULL
3     2001           9/3/2011     2
5     3001           9/18/2011    NULL

Answer 6

SELECT  ID,
        AccountNumber,
        Date,
        NextDate,
        DATEDIFF("D", Date, NextDate)
FROM    (   SELECT  ID, 
                    AccountNumber,
                    Date,
                    (   SELECT  MIN(Date) 
                        FROM    YourTable T2
                        WHERE   T2.Accountnumber = T1.AccountNumber
                        AND     T2.Date > T1.Date
                    ) AS NextDate
            FROM    YourTable T1
        ) AS T

连续行之间的日期差异

6 个答案: