连接一列中的多个值，从另一个表中选择

Question

鉴于这些简化的多项选择表，有时不止一个答案是正确的：

STUDENT_ANSWERS
AnswerID | StudentID | QuestionID | Answers
-------------------------------------------
       1 |         1 |          1 | C,D

QUESTION_ANSWERS
QuestionID | Answer | Text
-------------------------------------------------
         1 |      A | This is answer A
         1 |      B | B could also be correct
         1 |      C | Maybe it's C?
         1 |      D | Definitely D!

如何进行选择，将答案翻译成他们的描述？

我的开始：

SELECT *
FROM STUDENT_ANSWERS sa
LEFT OUTER JOIN QUESTION_ANSWERS qa ON qa.Answer IN sa.Answers???
 -- Doesn't seem to work as IN requires a format of ('C','D') while I have 'C,D'

期望的输出：

AnswerID | StudentID | QuestionID | AnswerDescriptions
-------------------------------------------
       1 |         1 |          1 | Maybe it's C?,Definitely D!

因此，描述只需替换代码，而不是为每个答案获取一行。

Answer 1

您的问题是表STUDENT_ANSWERS 的结构。每个答案应该有一行：

答案ID |学生证 |问题ID |回答 ------------------------------------------ 1 | 1 | 1 | C 2 | 1 | 1 | D

---

现在，假设你不能做任何改变（阅读：修复）这个，你可以通过附加逗号和使用 LIKE 来伪造它：

select *
from STUDENT_ANSWERS a
join QUESTION_ANSWERS q on ',' + a.Answers + ',' like '%,' + q.Answer + ',%'
    and a.QuestionID = q.QuestionID

SQL Fiddle demo

注意这一点假设您永远不会在QUESTION_ANSWERS.Answer 中拥有文本,。它也永远无法使用索引，所以它会比慢更慢。

如果您绝对必须在数据库中将其格式化为一行，您可以使用STUFF 和FOR XML PATH('') 技巧来连接结果行。

Answer 2

这是仅使用 T-SQL 语句的完整工作示例。我会建议您创建单独的函数来拆分返回行集的CSV。此外，如果您正在处理大量数据，您可能需要创建一个 CLR 函数来拆分值。看看this article（这里有你需要的一切）。

DECLARE @StudentAnswers TABLE
(
     [AnswerID] INT
    ,[StudentID] INT
    ,[QuestionID] INT
    ,[Answers] VARCHAR(256)
);

DECLARE @QuestionAnswers TABLE
(
     [QuestionID] INT
    ,[Answer] CHAR
    ,[Text] VARCHAR(256)
);

INSERT INTO @StudentAnswers ([AnswerID], [StudentID], [QuestionID], [Answers])
VALUES (1, 1, 1, 'C,D')
      ,(2, 2, 1, 'A');

 INSERT INTO @QuestionAnswers ([QuestionID], [Answer], [Text])
 VALUES  (1, 'A', 'This is answer A')
        ,(1, 'B', 'B could also be correct')
        ,(1, 'C', 'Maybe it''s C?')
        ,(1, 'D', 'Definitely D!');

SELECT SA.[AnswerID]
      ,SA.[StudentID]
      ,SA.[QuestionID]
      ,T.c.value('.', 'CHAR')
      ,QA.[Text]
FROM @StudentAnswers SA
CROSS APPLY 
(
    SELECT CAST('<i>' + REPLACE([Answers], ',', '</i><i>') + '</i>' AS XML) Answers
) DS
CROSS APPLY DS.Answers.nodes('i') T(c)
INNER JOIN @QuestionAnswers QA
    ON SA.[QuestionID] = QA.[QuestionID]
    AND T.c.value('.', 'CHAR') = QA.[Answer];

Answer 3

试试这个

select answerid,studentid,a.QuestionID,group_concat(b.text) from student_answers a left join QUESTION_ANSWERS b on b.questionid= a.questionid and  FIND_IN_SET(b.Answer, a.Answers)
group by a.questionid

确实有效。