我如何使返回表的列等于另一个表的行，像这样？

Question

我有 3 张类似的桌子

         Dudes
==========================
 id |    name   
==========================
  1 |  "Marshal Mathers"
--------------------------
  2 |  "Limp Biscuit"
--------------------------


        Questions 
===========================
 id |       qtext
===========================
 1  | "What's up, dog?"
---------------------------
 2  | "C# or C++?"
---------------------------

          Answers
=======================================================
 id |  dude_id  | question_id |       atext
=======================================================
  1 |      1       |       1     |    "Nothing much"
-------------------------------------------------------
  2 |      2       |       1     |     "Oh, you know"
-------------------------------------------------------
  3 |      1       |       2     |         "C++"
-------------------------------------------------------- 

我对 Make-A-Wish Foundation (Stack Overflow) 的希望是一个在该示例中返回的 sproc

===========================================================
          name         |  What's up, dog?  |  C# or C++?   
===========================================================
    "Marshal Mathers"  |   "Nothing much"  |    "C++"
-----------------------------------------------------------
     "Limp Biscuit"    |   "Oh, you know"  |     NULL
------------------------------------------------------------

这有可能吗？如果可以，该怎么做？

Answer 1

这是 sgeddes 答案的动态版本。这种技术被称为dynamic crosstab.

DECLARE @sql NVARCHAR(MAX)

SELECT @sql = 
'SELECT
    d.name' + CHAR(10)

SELECT @sql = @sql +
'   , MAX(CASE WHEN q.id = ' + CONVERT(VARCHAR(10), id) + ' THEN a.atext END) AS ' 
    + QUOTENAME(qtext) + CHAR(10)
FROM #Questions

SELECT @sql = @sql +
'FROM #Dudes d
LEFT JOIN #Answers a
    INNER JOIN #Questions q
        ON q.id = a.question_id
    ON a.dude_id = d.id
GROUP BY d.id, d.name
ORDER BY d.id'

PRINT(@sql)
EXEC(@sql)

DEMO

Answer 2

假设我正确理解您的问题，您可以使用conditional aggregation 到pivot 您的结果与outer join：

select d.id, d.name, 
       max(case when q.id = 1 then q.qtext end) whats_up_dog,
       max(case when q.id = 2 then q.qtext end) c_sharp_plus
from dudes d
       left join answers a on d.id = a.partner_id 
       left join questions q on q.id = a.question_id
group by d.id, d.name