连接两个表并返回Null记录

Question

我有两个表:Training表和Training_Log表

Create Table Trainings
(  Training_ID int
  ,Train_Title varchar(20)
 )

Create Table Training_Log
(   Log_ID int
  , EmployeeName varchar(20)
  , Training_ID int
  , Training_Date date
)
> insert into Trainings values(1,'Sales Training') 
  insert into Trainings values(2,'Risk Management') 
  insert into Trainings values(3,'First Aid')
> 
> insert into  Training_Log values (1, 'Bob Marley', 2, '10/01/2005')
> insert into  Training_Log values (2, 'Bob Marley', 3, '02/05/2006')
> insert into Training_Log values (3,'Ralph Kimball',1,'01/04/2004')

如果我想要获得如下结果集，我如何编写SQL：

   EmployeeName               Train_Title             Train_Date  
   Bob Marley                 Sales Training             Null  
   Bob Marley                 Risk Management           10/01/2005  
   Bob Marley                 First Aid                 02/05/2006
   Ralph Kimball              Sales Training            01/04/2004  
   Ralph Kimball              Risk Management            Null  
   Ralph Kimball              First Aid                  Null

Answer 1

我同意Brian Driscol的观点，你应该考虑为员工创建一个表格。

无论如何，此查询返回您想要的内容：

WITH AllTrainingCombinations AS 
(
    SELECT DISTINCT  EmployeeName, Train_Title, Trainings.Training_ID
    FROM Training_Log 
        CROSS JOIN Trainings
)

SELECT A.EmployeeName, A.Train_Title, T.Training_Date
FROM AllTrainingCombinations AS A
    LEFT JOIN Training_Log AS T ON A.EmployeeName = T.EmployeeName 
        AND A.Training_ID = T.Training_ID


Bob Marley  First Aid   2006-02-05
Bob Marley  Risk Management 2005-10-01
Bob Marley  Sales Training  NULL
Ralph Kimball   First Aid   NULL
Ralph Kimball   Risk Management NULL
Ralph Kimball   Sales Training  2004-01-04

Answer 2

这将会起作用，但它不会找到没有接受过至少一次培训日志的员工。

select e.EmployeeName, t.Train_Title, tl.Train_Date
from
    Trainings t cross join
    (select distinct EmployeeName from Training_Log) e
    left outer join
    Training_Log tl
        on tl.Training_ID = t.Training_ID and tl.EmployeeName = e.EmployeeName
order by e.EmployeeName, t.Training_ID

这与李嘉图的答案几乎相同，只是形式略有不同。