Sql 如何根据值建立与列的关系
4我有以下表格:Sql 如何根据值建立与列的关系,sql,ms-access,Sql,Ms Access,4我有以下表格: day_shift id_dshift dshift_name on_duty off_duty in_start in_end out_start out_end workday 1001 ds_normal 7:00 15:00 7:00 10:00 11:00 20:00 1 1002 ds
day_shift
id_dshift dshift_name on_duty off_duty in_start in_end out_start out_end workday
1001 ds_normal 7:00 15:00 7:00 10:00 11:00 20:00 1
1002 ds_Saturday 7:00 14:00 7:00 10:00 11:00 14:00 1
week_shift
id_wshift wshift_name mon tue wed thu fri sat sun
2001 ws_normal 1001 1001 1001 1001 1001 1002 1001
2002 ws_2013_w1 0 1001 1001 1001 1001 1001 0
2003 ws_2013_w2 1003 1001 1001 1001 1001 1002 1001
daily_attendance
emp_id checkdate in out emp_shift_id
10 15/06/2013 7:10 15:05 2001 <-- saturday
10 16/06/2013 7:05 15:03 2001 <-- sunday
在每日出勤的第一行,因为工作日是星期六,所以我想得到week_shift.sat(1002)的值
如果工作日是星期日,我想得到week_shift.sun(1001)的值
所以我从白班得到上下班值
如何在查询中执行此操作?这里的诀窍是在Access中创建一个名为[week\u shift\u transformed]的已保存查询,以将[week\u shift]表转换为一周中每一天的单独行:
SELECT id_wshift, wshift_name, 1 AS [weekday], [sun] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 2 AS [weekday], [mon] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 3 AS [weekday], [tue] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 4 AS [weekday], [wed] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 5 AS [weekday], [thu] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 6 AS [weekday], [fri] as id_dshift FROM week_shift
UNION ALL
SELECT id_wshift, wshift_name, 7 AS [weekday], [sat] as id_dshift FROM week_shift
那会给你
id_wshift wshift_name weekday id_dshift
--------- ----------- ------- ---------
2001 ws_normal 1 1001
2002 ws_2013_w1 1 0
2003 ws_2013_w2 1 1001
2001 ws_normal 2 1001
2002 ws_2013_w1 2 0
2003 ws_2013_w2 2 1003
2001 ws_normal 3 1001
2002 ws_2013_w1 3 1001
2003 ws_2013_w2 3 1001
2001 ws_normal 4 1001
2002 ws_2013_w1 4 1001
2003 ws_2013_w2 4 1001
2001 ws_normal 5 1001
2002 ws_2013_w1 5 1001
2003 ws_2013_w2 5 1001
2001 ws_normal 6 1001
2002 ws_2013_w1 6 1001
2003 ws_2013_w2 6 1001
2001 ws_normal 7 1002
2002 ws_2013_w1 7 1001
2003 ws_2013_w2 7 1002
然后可以使用如下查询:
SELECT da.emp_id, da.checkdate, da.in, da.out, ds.on_duty, ds.off_duty
FROM
daily_attendance da
INNER JOIN
(
week_shift_transformed wtt
INNER JOIN
day_shift ds
ON ds.id_dshift = wtt.id_dshift
)
ON wtt.weekday = Weekday(da.checkdate)
AND wtt.id_wshift = da.emp_shift_id
根据问题中的解释,结果关闭值不正确。在2013年6月15日(星期六)上,id_dshift=1002的is应为14:00,而对于2013年6月16日(星期日),id_dshift=1001的is应为15:00。观察速度很快。。我会编辑。。谢谢。考虑规范你的数据。表格不是电子表格。@草莓:我已经考虑过了,不是每周轮班1行,而是7行。如果我所要求的不能做到,我会这样做。是否能做到并不重要!
SELECT da.emp_id, da.checkdate, da.in, da.out, ds.on_duty, ds.off_duty
FROM
daily_attendance da
INNER JOIN
(
week_shift_transformed wtt
INNER JOIN
day_shift ds
ON ds.id_dshift = wtt.id_dshift
)
ON wtt.weekday = Weekday(da.checkdate)
AND wtt.id_wshift = da.emp_shift_id