django ORM的一些高级操作

假设有以下ORM模型:

from django.db import models

class Student(models.Model):
    """学生表"""
    name = models.CharField(max_length=100)
    gender = models.SmallIntegerField()

    class Meta:
        db_table = 'student'

class Course(models.Model):
    """课程表"""
    name = models.CharField(max_length=100)
    teacher = models.ForeignKey("Teacher",on_delete=models.SET_NULL,null=True)
    class Meta:
        db_table = 'course'

class Score(models.Model):
    """分数表"""
    student = models.ForeignKey("Student",on_delete=models.CASCADE)
    course = models.ForeignKey("Course",on_delete=models.CASCADE)
    number = models.FloatField()

    class Meta:
        db_table = 'score'

class Teacher(models.Model):
    """老师表"""
    name = models.CharField(max_length=100)

    class Meta:
        db_table = 'teacher'

使用之前学到过的操作实现下面的查询操作:

  1. 查询平均成绩大于60分的同学的id和平均成绩;

  2. 查询所有同学的id、姓名、选课的数量、总成绩;

  3. 查询姓“李”的老师的个数;

  4. 查询没学过“李老师”课的同学的id、姓名;

  5. 查询学过课程id为1和2的所有同学的id、姓名;

  6. 查询学过“黄老师”所教的“所有课”的同学的id、姓名;

  7. 查询所有课程成绩小于60分的同学的id和姓名;

  8. 查询没有学全所有课的同学的id、姓名;

  9. 查询所有学生的姓名、平均分,并且按照平均分从高到低排序;

  10. 查询各科成绩的最高和最低分,以如下形式显示:课程ID,课程名称,最高分,最低分;

  11. 查询没门课程的平均成绩,按照平均成绩进行排序;

  12. 统计总共有多少女生,多少男生;

  13. 将“黄老师”的每一门课程都在原来的基础之上加5分;

  14. 查询两门以上不及格的同学的id、姓名、以及不及格课程数;

  15. 查询每门课的选课人数;

参考答案:

  1. 查询平均成绩大于60分的同学的id和平均成绩;

    rows = Student.objects.annotate(avg=Avg("score__number")).filter(avg__gte=60).values("id","avg")
    for row in rows:
    print(row)
  2. 查询所有同学的id、姓名、选课的数、总成绩;

    rows = Student.objects.annotate(course_nums=Count("score__course"),total_score=Sum("score__number"))
    .values("id","name","course_nums","total_score")
    for row in rows:
    print(row)
  3. 查询姓“李”的老师的个数;

    teacher_nums = Teacher.objects.filter(name__startswith="李").count()
    print(teacher_nums)
  4. 查询没学过“黄老师”课的同学的id、姓名;

    rows = Student.objects.exclude(score__course__teacher__name="黄老师").values('id','name')
    for row in rows:
    print(row)
  5. 查询学过id为1和2的所有同学的id、姓名;

    rows = Student.objects.filter(score__course__in=[1,2]).distinct().values('id','name')
    for row in rows:
    print(row)
  6. 查询学过“黄老师”所教的所有课的同学的学号、姓名;

    rows = Student.objects.annotate(nums=Count("score__course",filter=Q(score__course__teacher__name='黄老师')))
    .filter(nums=Course.objects.filter(teacher__name='黄老师').count()).values('id','name')
    for row in rows:
    print(row)
  7. 查询所有课程成绩小于60分的同学的id和姓名;

    students = Student.objects.exclude(score__number__gt=60)
    for student in students:
    print(student)
  8. 查询没有学全所有课的同学的id、姓名;

    students = Student.objects.annotate(num=Count(F("score__course"))).filter(num__lt=Course.objects.count()).values('id','name')
    for student in students:
    print(student)
  9. 查询所有学生的姓名、平均分,并且按照平均分从高到低排序;

    students = Student.objects.annotate(avg=Avg("score__number")).order_by("-avg").values('name','avg')
    for student in students:
    print(student)
  10. 查询各科成绩的最高和最低分,以如下形式显示:课程ID,课程名称,最高分,最低分:

    courses = Course.objects.annotate(min=Min("score__number"),max=Max("score__number")).values("id",'name','min','max')
    for course in courses:
    print(course)
  11. 查询每门课程的平均成绩,按照平均成绩进行排序;

    courses = Course.objects.annotate(avg=Avg("score__number")).order_by('avg').values('id','name','avg')
    for course in courses:
    print(course)
  12. 统计总共有多少女生,多少男生;

    rows = Student.objects.aggregate(male_num=Count("gender",filter=Q(gender=1)),female_num=Count("gender",filter=Q(gender=2)))
    print(rows)
  13. 将“黄老师”的每一门课程都在原来的基础之上加5分;

    rows = Score.objects.filter(course__teacher__name='黄老师').update(number=F("number")+5)
    print(rows)
  14. 查询两门以上不及格的同学的id、姓名、以及不及格课程数;

    students = Student.objects.annotate(bad_count=Count("score__number",filter=Q(score__number__lt=60))).filter(bad_count__gte=2).values('id','name','bad_count')
    for student in students:
    print(student)
  15. 查询每门课的选课人数;

    courses = Course.objects.annotate(student_nums=Count("score__student")).values('id','name','student_nums')
    for course in courses:
    print(course)

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转载自blog.csdn.net/qq_37049050/article/details/82108056