假设有7个这样的表：
class_teacher_lesson
class_teacher_lession_ID, class_ID, teacher_ID, lesson_ID
类
class_ID, class
教师
teacher_ID, teacher
课
lesson_ID, lesson
学生
student_ID, class_ID, ..., ....
survey_answer
survey_answer_ID, survey_question_ID, class_teacher_lession_ID, survey_answer
survey_question
survey_question_ID, survey_question,
结果应该是这样的：
我将通过循环得到这个：
课程数学：
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%
课程英语：
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，我已经尝试过这个：
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
结果如下：
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?
答案 0 :(得分：0)
select survey_question, teacher_id, lesson_id ,
count(1) as Total_student_answer_survey,
(select count(1) from student) as total_student,
count(1)/(select count(1) from student) * 100 as survey_percentage
from survey_answer sa
inner join survey_question sq
on sa.survey_question_id = sq.survey_question_id
inner join class_teacher_lesson ctl
on sa.class_teacher_lesson_id = ctl.class_teacher_lesson_id
group by teacher_id, lesson_id, survey_question_id
答案 1 :(得分：0)
我假设您的选择中的enquete_vraag和您问题中的survey_question实际上是同一列，而这只是翻译中遗漏的内容（否则查询将无法选择列不在小组中。）
在这种情况下，您实际需要做的就是从group by子句中删除survey_answer.survey_answer（您也可以将一些左连接更改为内部，这可以提高性能。）。
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',
100.0 * COUNT(enquete_antwoord_ID) / COUNT(student_ID) AS 'survey_answer AVG'
FROM Teacher
INNER JOIN class_teacher_lesson
ON teacher.teacher_ID = class_teacher_lesson.teacher_ID
INNER JOIN lesson
ON class_teacher_lesson.class_ID = lesson.lesson_ID
INNER 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_question
然而（虽然这完全是主观的）我建议使用表别名清理查询，也不建议使用单引号来识别列别名（以避免与文字混淆），并且（这更加主观）我添加了SQL-Server特定的Alias = Column_name语法，我个人认为这种语法更具可读性（Perhaps some more eloquent reasons to switch are presented here by Aaron Bertrand）。
SELECT t.teacher,
c.class,
l.lesson,
[Number of students] = COUNT(s.student_ID),
[Total students answered survey] = COUNT(sa.enquete_antwoord_ID),
[Survey Question] = sq.enquete_vraag,
[% Students Answered] = 100.0 * COUNT(sa.enquete_antwoord_ID) / COUNT(s.student_ID)
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.lesson_ID
INNER JOIN class c
ON ctl.class_ID = c.class_ID
LEFT JOIN student s
ON c.class_ID = s.class_ID
LEFT JOIN survey_answer sa
ON ctl.class_teacher_lesson_ID = sa.class_teacher_lesson
LEFT JOIN survey_question sq
ON sa.survey_question_ID = sq.survey_question_ID
WHERE t.teacher_ID = '1158'
AND l.lesson_ID = '449'
GROUP BY c.class, l.lesson, t.teacher, sq.survey_question;
这两种解决方案都应该给出相同的结果，我并不想强迫我对任何人的偏好，我只是提出选择。对于您发现哪些更具可读性，这只是一个偏好问题。
修改
我认为由于将多个一对多关系加入到相同的“一个”中，您正在进行交叉加入，这导致了零调查问题。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';