SQL-联接中的条件列选择

我不确定是否可以使用TSQL实现此场景。我有一个叫做WorkingDays的表格，里面有这个信息

ID  |  EmployeeId |  Monday | Tuesday | Wednesday | Thursday | Friday
----------------------------------------------------------------------
1   | 1           | 2       | 2       | 3         | 6        | 5
2   | 2           | 1       | 7       | 5         | 2        | 3

days列存储WorkingSchedule表的ID，该表有以下列：

ID          int    Primary Key
StartTime   time
EndTime     time

因此，我需要根据当前日期获取员工的开始时间和结束时间

我需要从查询中得到的是开始和结束时间，具体取决于日期。我要筛选的日期是使用getdate函数反当前日期

因此，需要选择正确的日期列名来进行联接

---

如何实现此方案？

动态sql版本：

declare @sql nvarchar(max) ='
select 
    t.EmployeeId
  , StarTime = max(case when t.rn=1 then '+quotename(datename(weekday,getdate()))+' end)
  , EndTime  = max(case when t.rn=2 then '+quotename(datename(weekday,getdate()))+' end)
from (
  select *
    , rn = row_number() over (partition by t.EmployeeId order by t.Id)
    from t
    ) t
group by t.EmployeeId;'
exec sp_executesql @sql;

rextester演示：

返回：

+------------+----------+---------+
| EmployeeId | StarTime | EndTime |
+------------+----------+---------+
|          1 |        5 |       3 |
+------------+----------+---------+

+------------+---------+----------+---------+
| EmployeeId | WorkDay | StarTime | EndTime |
+------------+---------+----------+---------+
|          1 | Friday  |        5 |       3 |
+------------+---------+----------+---------+

+----+------------+---------+-----------+------+
| Id | EmployeeId | WorkDay | StartEnd  | Time |
+----+------------+---------+-----------+------+
|  1 |          1 | Friday  | StartTime |    5 |
|  2 |          1 | Friday  | EndTime   |    3 |
+----+------------+---------+-----------+------+

根据您希望输出的方式，这里有两种不使用动态sql的其他方式：

两者都用于取消打印数据，WorkDay=datenameweekday，getdate用于获取当前WorkDay列

对于单行输出，我们添加了一些条件聚合：

/* one row per employeeId */
select 
    t.EmployeeId
  , x.WorkDay
  , StarTime = max(case when t.rn=1 then x.Time end)
  , EndTime  = max(case when t.rn=2 then x.Time end)
from (
  select *
    , rn = row_number() over (partition by t.EmployeeId order by t.Id)
    from t
    ) t
  cross apply (values 
    ('Monday',Monday),('Tuesday',Tuesday),('Wednesday',Wednesday)
    ,('Thursday',Thursday),('Friday',Friday)
  ) x (WorkDay,Time)
where WorkDay = datename(weekday,getdate())
group by t.EmployeeId, x.WorkDay

返回：

+------------+----------+---------+
| EmployeeId | StarTime | EndTime |
+------------+----------+---------+
|          1 |        5 |       3 |
+------------+----------+---------+

+------------+---------+----------+---------+
| EmployeeId | WorkDay | StarTime | EndTime |
+------------+---------+----------+---------+
|          1 | Friday  |        5 |       3 |
+------------+---------+----------+---------+

+----+------------+---------+-----------+------+
| Id | EmployeeId | WorkDay | StartEnd  | Time |
+----+------------+---------+-----------+------+
|  1 |          1 | Friday  | StartTime |    5 |
|  2 |          1 | Friday  | EndTime   |    3 |
+----+------------+---------+-----------+------+

如果希望输出为两行，如当前输出：

/* two rows per employeeId */
select 
    t.Id
  , t.EmployeeId
  , x.WorkDay
  , t.StartEnd
  , x.Time
from (
  select *
    , StartEnd = case 
        when row_number() over (partition by t.EmployeeId order by t.Id) = 1 
          then 'StartTime' 
        else 'EndTime' 
        end 
    from t
    ) t
  cross apply (values 
    ('Monday',Monday),('Tuesday',Tuesday),('Wednesday',Wednesday)
    ,('Thursday',Thursday),('Friday',Friday)
  ) x (WorkDay,Time)
where WorkDay = datename(weekday,getdate());

返回：

+------------+----------+---------+
| EmployeeId | StarTime | EndTime |
+------------+----------+---------+
|          1 |        5 |       3 |
+------------+----------+---------+

+------------+---------+----------+---------+
| EmployeeId | WorkDay | StarTime | EndTime |
+------------+---------+----------+---------+
|          1 | Friday  |        5 |       3 |
+------------+---------+----------+---------+

+----+------------+---------+-----------+------+
| Id | EmployeeId | WorkDay | StartEnd  | Time |
+----+------------+---------+-----------+------+
|  1 |          1 | Friday  | StartTime |    5 |
|  2 |          1 | Friday  | EndTime   |    3 |
+----+------------+---------+-----------+------+

提示：datenameweekday，getdate返回当前区域设置中的工作日名称！这可能更好：

select wd.Employee, ws.StartTime, ws.EndTime
  from WorkingDays wd
  join WorkingSchedule ws on ws.Id = case datepart(weekday, getdate())
                                      when 1 then wd.Monday
                                      when 2 then wd.Tuesday
                                      when 3 then wd.Wednesday
                                      when 4 then wd.Thursday
                                      when 5 then wd.Friday
                                      else        0
                                     end

---

但是，根据您的设置，您必须检查第0周、第1周的第一天。

如果您可以根据上述数据提供示例输出，这将非常有用。workingdays表中的id是什么。如果是那个表id，那么同一个员工怎么可能包含两个不同的工作计划id，我是说同一个emp id有两行，每行只有一个员工！每个员工在每个工作日都有工作时间。所以问题很清楚。我们也不需要一个示例输出，因为问题是关于连接的方式，不是确切的结果。就我所知：好吧，我把这个问题读了好几遍，仍然没有任何线索+1