如何对多个表使用SQL联接操作

假设有7个这样的表：

Teacher:  Class:   Lesson:  Number of students:  Filled in:    Survey_question:
t1        c1       math     90                   54            question1?

班级教师课程

class_teacher_lession_ID, class_ID, teacher_ID, lesson_ID

lesson_ID, lesson

survey_question_ID, survey_question, 

课程

class_ID, class

老师

teacher_ID, teacher

survey_answer_ID, survey_question_ID, class_teacher_lession_ID, survey_answer

课程

class_teacher_lession_ID, class_ID, teacher_ID, lesson_ID

lesson_ID, lesson

survey_question_ID, survey_question, 

学生

student_ID, class_ID, ..., ....

调查\u答案

teacher_ID, teacher

survey_answer_ID, survey_question_ID, class_teacher_lession_ID, survey_answer

调查问题

class_teacher_lession_ID, class_ID, teacher_ID, lesson_ID

lesson_ID, lesson

survey_question_ID, survey_question, 

结果应该是这样的：

Teacher:  Class:   Lesson:  Number of students:  Filled in:    Survey_question:
t1        c1       math     90                   54            question1?

我将通过一个循环得到这个：

数学课：

survey_question (where teacher_id = 11 and lesson_id = 13):
Total Students            Total Students Answered Survey         survey_answer AVG
     155                                45                              79%

survey_question 2 (where teacher_id = 11 and lesson_id = 13):
Total Students            Total Students Answered Survey         survey_answer AVG
     155                                99                              87%

第二课英语：

survey_question (where teacher_id = 11 and lesson_id = 15):
Total Students            Total Students Answered Survey         survey_answer AVG
     90                                20                              50%

survey_question 2 (where teacher_id = 11 and lesson_id = 15):
Total Students            Total Students Answered Survey         survey_answer AVG
     90                                25                              34%

到目前为止，我已经尝试过：

SELECT teacher AS 'Teacher', class AS 'Class', lesson AS 'Lesson',
COUNT(student_ID) AS 'Number of students',
COUNT(enquete_antwoord_ID) AS 'Total students answered survey', 
enquete_vraag AS 'Survey Question'
FROM Teacher
LEFT JOIN class_teacher_lesson
ON teacher.teacher_ID = class_teacher_lesson.teacher_ID
LEFT JOIN lesson
ON class_teacher_lesson.class_ID = lesson.lesson_ID
LEFT JOIN class
ON class_teacher_lesson.class_ID = class.class_ID
LEFT JOIN student
ON class.class_ID = student.class_ID
LEFT JOIN survey_answer
ON class_teacher_lesson.class_teacher_lesson_ID = 
survey_answer.class_teacher_lesson
LEFT JOIN survey_question
ON survey_answer.survey_question_ID = survey_question.survey_question_ID
WHERE teacher.teacher_ID = '1158' AND lesson.lesson_ID = '449'
GROUP BY class.class, lesson.lesson, teacher.teacher,
survey_answer.survey_answer, survey_question

根据这一结果：

Teacher:  Class:   Lesson:  Number of students:  Filled in:    Survey_question:
t1        c1       math     54                   54            question1?
t1        c1       math     24                   0             NULL
t1        c1       math     22                   0             NULL

结果应该是这样的：

Teacher:  Class:   Lesson:  Number of students:  Filled in:    Survey_question:
t1        c1       math     90                   54            question1?

---

我假设您的选择中的

enquete\u vraag

和您的问题中的

survey\u question

实际上是同一列，这只是翻译中遗漏的内容（否则查询将无法选择不在group by中的列）

在这种情况下，您实际上需要做的就是从GROUPBY子句中删除

survey\u-answer.survey\u-answer

（您还可以将一些左侧联接更改为INTERNAR，这可以提高性能）

但是（尽管这完全是主观的）我建议使用表别名清理查询，也不要使用单引号来标识列别名（以避免与文字混淆），而且（这更主观）我添加了SQL Server特有的

Alias=column\u name

语法，我个人认为这更具可读性（）

这两种解决方案应该得到相同的结果，我并不是想把我的偏好强加给任何人，我只是提出一些选择。这只是一个偏好的问题，你会发现哪个更具可读性

编辑

我认为，由于将多个“一-多”关系连接到同一个“一”，这导致了无效的调查问题，因此您正在获得交叉连接。我认为将聚合移动到子查询可能会有所帮助：

SELECT  t.teacher, 
        c.class, 
        l.lesson,
        [Number of students] = ISNULL(s.Students, 0),
        [Total students answered survey] = ISNULL(sq.Answers, 0), 
        sq.SurveyQuestion,
        [% Students Answered] = 100.0 * ISNULL(sq.Answers, 0) / NULLIF(s.Students, 0)
FROM    Teacher t
        INNER JOIN class_teacher_lesson ctl
            ON t.teacher_ID = ctl.teacher_ID
        INNER JOIN lesson l
            ON ctl.class_ID = l.class_ID
        INNER JOIN class c
            ON ctl.class_ID = c.class_ID
        LEFT JOIN 
        (   SELECT  Class_ID, Students = COUNT(*)
            FROM    student s
            GROUP BY Class_ID
        ) S
            ON c.class_ID = s.class_ID
        LEFT JOIN 
        (   SELECT  sa.class_teacher_lesson_ID, 
                    sq.SurveyQuestion,
                    Answers = COUNT(*)
            FROM    survey_answer sa
                    LEFT JOIN survey_question sq
                        ON sa.survey_question_ID = sq.survey_question_ID
            GROUP BY sa.class_teacher_lesson_ID, sq.SurveyQuestion
        ) sq
            ON ctl.class_teacher_lesson_ID = sq.class_teacher_lesson_ID
WHERE   t.teacher_ID = '1158' 
AND     l.lesson_ID = '449';

---

请写下你迄今为止所做的努力。您可以在

Sql

查询中使用带有适当

Where

条件的

Join

操作。可能对你有帮助。或者试着读一下，你真正的问题是什么？您是否尝试过某项操作并遇到错误？你有没有尝试过一些东西却没有得到你期望的结果？你甚至不知道从哪里开始？我知道标题中有一个问题，但实际上标题应该表明您的一般问题，您的问题需要详细说明您的问题的细节。目前，您可能得到的答案只是对您面临的问题的猜测，而不是实际解决您的问题，并（希望）向您展示如何在将来解决类似问题。@wxyz I更新了question@KishanSarsechaGajjar我更新了问题MSG 209，16级，状态1，第19行不明确的列名“enquete_vraag_ID”。Msg 207，第16级，州1，第19行无效列名“总计学生答案调查”。Msg 207，级别16，状态1，第19行无效列名“total_student”。在SQL Server中，您不能通过别名引用选择列表中的表达式，必须重复该表达式。已删除别名。你能检查一下更新的oneNice jats吗，仍然有第一个错误：Msg 209，16级，状态1，第40行模棱两可的列名“survey\u question\u id”你可以在分组中添加sq.survey\u question\u id..抱歉遗漏了一些小东西谢谢你的回复不幸的是，我没有收到任何没有错误的回复msg.ON ctl.class\u id=l.lesson\u id=l.Lessure\u id内部连接类c ON ctl.class\u id=c.class\u id这是错误的，但我仍然没有得到正确的回复结果5行事实上，我从GROUPBY和select中删除了类，但现在我在这里得到了两行输出：您使用的是什么DBMS？这个问题被标记为SQL Server和MySQL，现在两者都没有。我使用SQL Server作为DBMS，而且学生填写的必须是1，因为在调查的答案中只有1行。