2134

Posted 2021-01-31 bigc008

 

　　

    INSTALLED_APPS = [
        ‘django.contrib.admin‘,
        ‘django.contrib.auth‘,
        ‘django.contrib.contenttypes‘,
        ‘django.contrib.sessions‘,
        ‘django.contrib.messages‘,
        ‘django.contrib.staticfiles‘,
        ‘web.apps.WebConfig‘,
        ‘rbac.apps.RbacConfig‘,
    ]

 

from django.db import models


class Menu(models.Model):
    """
    菜单表  一级菜单
    """
    title = models.CharField(max_length=32)
    icon = models.CharField(max_length=64, null=True, blank=True, verbose_name=‘图标‘)
    # 默认值为1
    weight = models.IntegerField(default=1, verbose_name=‘显示权重‘)

    def __str__(self):
        return self.title


class Permission(models.Model):
    """
    权限表
    可以做二级菜单的权限   menu 关联 菜单表
    不可以做菜单的权限    menu=null
    """
    url = models.CharField(max_length=32, verbose_name=‘权限‘)
    title = models.CharField(max_length=32, verbose_name=‘标题‘)
    menu = models.ForeignKey(‘Menu‘, null=True, blank=True)
    # 自己关联自己
    parent = models.ForeignKey(‘self‘, null=True, blank=True)
    # 写法二
    # parent = models.ForeignKey(‘Permission‘, null=True, blank=True)
    name = models.CharField(max_length=32, verbose_name=‘URL别名‘)

    class Meta:
        verbose_name_plural = ‘权限‘
        verbose_name = ‘权限‘

    def __str__(self):
        return self.title


class Role(models.Model):
    """
    角色表
    """
    name = models.CharField(max_length=32, verbose_name=‘名称‘)
    permissions = models.ManyToManyField(‘Permission‘, verbose_name=‘角色拥有的权限‘, blank=True)

    def __str__(self):
        return self.name


class User(models.Model):
    """
    用户表
    """
    name = models.CharField(max_length=32, verbose_name=‘名称‘)
    password = models.CharField(max_length=32, verbose_name=‘密码‘)
    roles = models.ManyToManyField(‘Role‘, verbose_name=‘用户拥有的角色‘, blank=True)

    def __str__(self):
        return self.name