從一個列中加入多個值，從另一個表中選擇

Question

鑒於這些簡化的多選表有時不止一個答案是正確的：

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!

如何進行選擇以轉換其描述的答案？

我的開始：

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'

期望的輸出：

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

因此，描述只需替換代碼而不是為每個答案獲取單行。

Answer 1

你的問題是表STUDENT_ANSWERS的結構。 每個答案應該有一行：

AnswerID | StudentID | QuestionID | Answer
-------------------------------------------
       1 |         1 |          1 | C
       2 |         1 |          1 | D

---

現在，假設你無法做任何改變（閱讀：修復）這個，你可以通過附加逗號和使用LIKE來捏造它：

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

SQL小提琴演示

請注意這是假設你永遠不會有文本,在QUESTION_ANSWERS.Answer 。 它也永遠不能使用索引，所以它會比慢速慢。

如果您絕對必須將數據庫中的格式設置為一行，則可以使用STUFF和FOR XML PATH('')技巧來連接生成的行。

Answer 2

這是僅使用T-SQL語句的完整工作示例。 我建議您創建單獨的函數來拆分返回行集的CSV 。 此外，如果您正在處理大量數據，則可能需要創建用於拆分值的CLR函數。 看看這篇文章 （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

嘗試這個

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

絕對有效。