Question

我正在尝试为网站设计数据库结构，以便人们打开个人资料。我试图弄清楚我选择的设计是否足够好。我怀疑它可能有问题的原因是我在许多表之间使用了很多关系。这样，当一个人的页面被加载时，幕后有许多JOIN，这肯定会造成瓶颈。如果你能帮助我弄清楚设计是否合适或我是否应该重新考虑它，我将不胜感激。

所以我想出了以下设计：

class Person(models.Model):

    GENDER_CHOICES = (
                    ('M', 'Male'),
                    ('F', 'Female'),
    )

    name = models.CharField(max_length=200)
    # Each person can have one profession
    profession = models.ForeignKey(Profession)
    email = models.CharField(max_length=100, blank=True)
    website = models.CharField(max_length=200, blank=True)
    gender = models.CharField(max_length=1, choices=GENDER_CHOICES))
    birth_date = models.DateField(blank=True, null=True)    

class Profession(models.Model):    
    type = models.CharField(max_length=20, blank=False, choices = ("Teacher","Pilot","Politician"))

#Each person can have several geolocations (represent the person's home/office/other address)
class Geolocation(models.Model):    
    latitude = models.FloatField()
    longitude = models.FloatField()
    address = models.TextField()
    related_person = models.ForeignKey(Person, blank=False, null=False)

#Each person can have several medias (you tube movies)
class Media(models.Model):
    youtube_id = models.CharField(max_length=100)
    related_person = models.ForeignKey(Person, blank=False, null=False)

#Each person can have several websites
class Websites(models.Model):
    website_url = models.CharField(max_length=200, blank=False)
    website_name = models.CharField(max_length=200)
    related_person = models.ForeignKey(Person, blank=False, null=False)
    #See below...
    related_link_farm = models.ForeignKey(LinkFarms, blank=True, null=True)

#Each website that refers to a person must be also refered to a link farm
class LinkFarms(models.Model):
    farm_name = models.CharField(max_length=200, blank=False)

请注意，对于每个加载的页面，至少有5个表连接。

谢谢，

梅尔

Answer 1

可以规范化数据。这就是关系数据库的用途！你的设计看起来很好。如果它成为一个问题，你总是可以缓存页面渲染。

Answer 2

也许你可以删除这样的职业表：

class Person(models.Model):
    ...

    TEACHER_ID = 'te'
    PILOT_ID = 'pi'
    POLITITAN_ID = 'po'
    PROFESSION_CHOICES = (
        (TEACHER_ID, 'Teacher'),
        (PILOT_ID, 'Pilot'),
        (POLITITAN_ID, 'Polititian')
    )
    ...
    profession = db.CharField(max_length=2, choices=PROFESSION_CHOICES)

然后你可以像这样查询所有的政治家：

polititans = Person.objects.filter(profession=Person.POLITITAN_ID)

使用Django设计数据库结构

2 个答案: