Question

如何将数据从table1插入到table2我正在使用MS Access。这就是我想要的。

表1

|id|    date    |  time |
| 3|  7/22/2017 |  8:00 |
| 3|  7/22/2017 | 17:00 |
| 4|  7/22/2017 |  8:00 |
| 4|  7/22/2017 | 12:00 |
| 4|  7/22/2017 | 13:00 |
| 4|  7/22/2017 | 17:00 |

我想在新桌子上这样做表2

|id|working date|time-in|time-out|time-in|time-out|
| 3|  7/22/2017 |  8:00 |        |       | 17:00  |
| 4|  7/22/2017 |  8:00 | 12:00  | 13:00 | 17:00  |

是否可以循环或这是一个查询？请帮忙

Answer 1

从执行不等自连接的查询开始，将顺序值分配给按[id]和[date]分组的行

SELECT 
    t1.id,
    t1.date,
    t1.time,
    COUNT(*) AS seq
FROM
    Table1 t1
    INNER JOIN
    Table1 t2
        ON t1.id = t2.id AND t1.date = t2.date AND t1.time >= t2.time
GROUP BY t1.id, t1.date, t1.time

给了我们

id  date        time      seq
--  ----------  --------  ---
 3  2017-07-22  08:00:00    1
 3  2017-07-22  17:00:00    2
 4  2017-07-22  08:00:00    1
 4  2017-07-22  12:00:00    2
 4  2017-07-22  13:00:00    3
 4  2017-07-22  17:00:00    4

在计算＆＃34;标记＆＃34;的查询中包裹它表示＆＃39; in1＆＃39;，＆＃39; out1＆＃39;，＆＃39; in2＆＃39;，＆＃39; out2＆＃39;等

SELECT
    [id],
    [date],
    [time],
    IIf([seq] Mod 2 = 1, "in", "out") & ([seq] + 1) \ 2 AS tag
FROM
(
    SELECT 
        t1.id,
        t1.date,
        t1.time,
        COUNT(*) AS seq
    FROM
        Table1 t1
        INNER JOIN
        Table1 t2
            ON t1.id = t2.id AND t1.date = t2.date AND t1.time >= t2.time
    GROUP BY t1.id, t1.date, t1.time
)

制造

id  date        time      tag 
--  ----------  --------  ----
 3  2017-07-22  08:00:00  in1 
 3  2017-07-22  17:00:00  out1
 4  2017-07-22  08:00:00  in1 
 4  2017-07-22  12:00:00  out1
 4  2017-07-22  13:00:00  in2 
 4  2017-07-22  17:00:00  out2

然后将整个事物包装在交叉表查询中

TRANSFORM Min([time]) AS MinOftime
SELECT [id], [date]
FROM
(
    SELECT
        [id],
        [date],
        [time],
        IIf([seq] Mod 2 = 1, "in", "out") & ([seq] + 1) \ 2 AS tag
    FROM
    (
        SELECT 
            t1.id,
            t1.date,
            t1.time,
            COUNT(*) AS seq
        FROM
            Table1 t1
            INNER JOIN
            Table1 t2
                ON t1.id = t2.id AND t1.date = t2.date AND t1.time >= t2.time
        GROUP BY t1.id, t1.date, t1.time
    )
)
GROUP BY [id], [date]
PIVOT [tag]

获得最终结果

id  date        in1       in2       out1      out2    
--  ----------  --------  --------  --------  --------
 3  2017-07-22  08:00:00            17:00:00          
 4  2017-07-22  08:00:00  13:00:00  12:00:00  17:00:00

确定各个交易记录的间隔的开始/结束时间

1 个答案: