将3个mysql查询组合成一个查询

我有一个学校管理系统，存储学生成绩并生成学生报告

一个学生要通过考试，他必须具备

平均60%或以上

英语成绩达到60%或以上

在包括英语在内的5门学科中获得至少60%的成绩

我有查询来计算最好的5个科目，并得出一个平均值

但是我需要我的查询能够检查通过科目的值，并且计算学生通过的科目数（包括英语），并在一个查询中显示该信息

SELECT student_id, round((SUM(t.mark))/5) average_mark from (
            select marks.student_id,  ROUND(AVG(mark)) as mark  from marks
                INNER JOIN teaching_loads ON teaching_loads.id=marks.teaching_load_id
                INNER JOIN subjects ON subjects.id=teaching_loads.subject_id
            where marks.student_id = "520" AND marks.assessement_id=1  
            GROUP BY subject_id
            order by (subject_id =2) desc, mark desc
            
            LIMIT 5
            
            )t ORDER BY round((SUM(t.mark))/5) DESC

如何构建一个查询来检查通过科目的值，并计算学生通过的科目数（包括英语），并在一个查询中显示该信息

差不多

学号：89 及格科目：6 及格科目分数：60分

在一个查询中，我希望能够获得所有这些数据，我该如何进行， 请帮帮我

下面是存储学生数据/分数及其相关表的数据库模式

分数表存储学生分数

CREATE TABLE `marks` (
  `id` bigint(20) UNSIGNED NOT NULL,
  `teacher_id` bigint(20) UNSIGNED NOT NULL,
  `student_id` bigint(20) UNSIGNED NOT NULL,
  `teaching_load_id` bigint(20) UNSIGNED NOT NULL,
  `assessement_id` bigint(20) UNSIGNED NOT NULL,
  `mark` int(11) NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

CREATE TABLE `teaching_loads` (
  `id` bigint(20) UNSIGNED NOT NULL,
  `teacher_id` bigint(20) UNSIGNED NOT NULL,
  `subject_id` bigint(20) UNSIGNED NOT NULL,
  `class_id` bigint(20) UNSIGNED NOT NULL,
  `session_id` bigint(20) UNSIGNED NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

教学负荷

CREATE TABLE `marks` (
  `id` bigint(20) UNSIGNED NOT NULL,
  `teacher_id` bigint(20) UNSIGNED NOT NULL,
  `student_id` bigint(20) UNSIGNED NOT NULL,
  `teaching_load_id` bigint(20) UNSIGNED NOT NULL,
  `assessement_id` bigint(20) UNSIGNED NOT NULL,
  `mark` int(11) NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

CREATE TABLE `teaching_loads` (
  `id` bigint(20) UNSIGNED NOT NULL,
  `teacher_id` bigint(20) UNSIGNED NOT NULL,
  `subject_id` bigint(20) UNSIGNED NOT NULL,
  `class_id` bigint(20) UNSIGNED NOT NULL,
  `session_id` bigint(20) UNSIGNED NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

主题表

CREATE TABLE `subjects` (
  `id` bigint(20) UNSIGNED NOT NULL,
  `subject_name` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL,
  `subject_type` enum('core','elective','non-value','passing_subject') COLLATE utf8mb4_unicode_ci NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

---

由于最后一个条件，这相当棘手。因此，我认为窗口功能：

select student_id,
       (case when avg(mark) >= 0.6 then 'Pass'
             when avg(case when subject = 'English' then mark end) > 0.6 then 'Pass'
             when avg(case when seqnum = 5 then mark end) >= 0.6 then 'Pass'
             else 'Fail'
         end) as pass_fail
from (select m.*, subject,
             row_number() over (partition by student_id order by (subject = 'English') desc, mark desc) as seqnum
      from marks m join
           teaching_loads tl
           on tl.id = m.teaching_load_id join
           subjects s
           on s.id = tl.subject_id
     ) ms
group by student_id;

这行吗

    select student_id,
       (case when avg(mark) >= 0.6 then 'Pass'
         when avg(case when subject = 'English' then mark end) > 0.6 then 'Pass'
         when avg(case when seqnum = 5 then mark end) >= 0.6 then 'Pass'
         else 'Fail'
     end) as pass_fail
       from (select m.*, subject,
         row_number() over (partition by student_id order by (subject = 'English') 
         desc, mark desc) as seqnum
          from marks m join
           teaching_loads tl
             on tl.id = m.teaching_load_id join
                subjects s
                  on s.id = tl.subject_id 
                   join
                    average_mark a
                    on a.student_id = marks.student_id
                    where a.average_mark > 60
            ) ms
   group by student_id;

---

我非常感谢你的帮助。我刚刚运行了这个查询，它确实有效，唯一的例外是它显示所有学生都通过了pass\u fail列，即使是那些不符合条件的学生。而且，查询的列仅限于student\u id和pass\u fail列。在这个用例中，我们还需要显示通过科目的分数和学生通过的科目数。