Question

我不确定如何准确地说出这一点，因为我刚刚启动了SQL并且刚刚完成了简单的连接。

首先从我的桌子开始：

tblQuestions

| Survey ID | User ID | Entry Date | Question 1 | Question 2 | Question 3 | Question 4 |
|     1     |   305   |    date    |      0     |      1     |     2      |      3     |

tblValues

|  VID  | TextValue |
|   0   |   Never   |
|   1   |   Rarely  |
|   2   | Sometimes |
|   3   |   Often   |

我想要一个SQL查询，它输出每个问题VID的TextValues。我无法直接存储文本值，因为数据将导出为图表。此外，我从我所理解的重复数据是糟糕的表设计和规则何时规范化表。

这可以通过单个查询吗？

编辑：我正在使用MS Access。期望的输出：

| Survey ID | User ID | Entry Date | Question 1 | Question 2 | Question 3 | Question 4 |
|     1     |   305   |    date    |    Never   |   Rarely   |  Sometimes |    Often   |

Answer 1

您想要将一个调查记录与四个答案记录连接起来，因此您必须四次查询答案表。使用表别名来区分四个记录：

select 
  q.survey_id, q.user_id, q.entry_date,
  v1.textvalue, v2.textvalue, v3.textvalue, v4.textvalue
from tblQuestions q
join tblValues v1 on v1.vid = q.question1
join tblValues v2 on v2.vid = q.question2
join tblValues v3 on v3.vid = q.question3
join tblValues v4 on v4.vid = q.question4;

更新：正如克里斯指出的那样，Access还有一些额外的加入要求。因此，上述标准SQL查询必须进行修改：

select 
  q.survey_id, q.user_id, q.entry_date,
  v1.textvalue, v2.textvalue, v3.textvalue, v4.textvalue
from (((tblQuestions q
inner join tblValues as v1 on v1.vid = q.question1)
inner join tblValues as v2 on v2.vid = q.question2)
inner join tblValues as v3 on v3.vid = q.question3)
inner join tblValues as v4 on v4.vid = q.question4;

Answer 2

就像一个开始：

CREATE TABLE Survey(ID INT NOT NULL IDENTITY CONSTRAINT PK_Survey PRIMARY KEY
                   ,BeginOfSurvey DATETIME NOT NULL);
CREATE TABLE Value(ID INT NOT NULL IDENTITY CONSTRAINT PK_Value PRIMARY KEY
                  ,Caption VARCHAR(100) NOT NULL);
CREATE TABLE QuestionText(ID INT NOT NULL IDENTITY CONSTRAINT PK_QuestionText PRIMARY KEY
                     ,Caption VARCHAR(100) NOT NULL);
CREATE TABLE Question(ID INT NOT NULL IDENTITY CONSTRAINT PK_Question PRIMARY KEY
                     ,SurveyID INT NOT NULL CONSTRAINT FK_Question_SurveyID FOREIGN KEY REFERENCES Survey(ID)
                     ,QuestionTextID INT NOT NULL CONSTRAINT FK_Question_QuestionTextID FOREIGN KEY REFERENCES QuestionText(ID)
                     ,ValueID INT NOT NULL CONSTRAINT FK_Question_ValueID FOREIGN KEY REFERENCES Value(ID));
INSERT INTO Value VALUES
 ('Never')
,('Rarely')
,('Sometimes')
,('Often');

INSERT INTO QuestionText VALUES
 ('Do you smoke?') 
,('Do you drink  alcohol?'); 

INSERT INTO Survey VALUES(GETDATE());

INSERT INTO Question VALUES
 (1,1,2) 
,(1,2,3);

SELECT *
FROM Survey AS s
INNER JOIN Question AS q ON s.ID=q.SurveyID
INNER JOIN QuestionText AS qt ON qt.ID=q.QuestionTextID
INNER JOIN Value AS v ON v.ID=q.ValueID

数据库查找另一个数据库查找的每个值

2 个答案: