Django动态渲染多层菜单

为后续给菜单设置权限管理方便，通过给页面模版菜单动态渲染，通过数据菜单表进行匹配需要渲染的菜单

 #Django表结构

 class Menus(models.Model):

     name = models.CharField(max_length=32, verbose_name=u'菜单名')
     parent = models.ForeignKey('self',
                                verbose_name=u'父级菜单',
                                null=True,
                                blank=True,
                                default='',
                                help_text=u'如果添加的是子菜单，请选择父菜单')
     show = models.BooleanField(verbose_name=u'是否显示',
                                default=False,
                                help_text=u'菜单是否显示，默认添加不显示')
     url = models.CharField(max_length=300,
                            verbose_name=u'菜单url地址',
                            null=True,
                            blank=True,
                            default='javascript:void(0)',
                            help_text=u'是否给菜单设置一个url地址')
     priority = models.IntegerField(verbose_name=u'显示优先级',
                                    null=True,
                                    blank=True,
                                    default=-1,
                                    help_text=u'菜单的显示顺序，优先级越大显示越靠前')
     permission_id = models.IntegerField(verbose_name=u'权限编号',
                                         help_text=u'给菜单设置一个编号，用于权限控制',
                                         error_messages={'field-permission_id': u'只能输入数字'})

     def __str__(self):
         return "{parent}{name}".format(name=self.name, parent="%s-->" % self.parent.name if self.parent else '')

     class Meta:
         verbose_name = u"菜单"
         verbose_name_plural = u"菜单"
         ordering = ["-priority", "id"]

 #admin.py

 from django.contrib import admin

 # Register your models here.

 from cloud.api import models as api_models

 class MenusAdmin(admin.ModelAdmin):
     ordering = ('-parent',)
     list_filter = ('name',)
     list_display = ['name', 'parent', 'show', 'url', 'priority', 'permission_id']
     fields = ['name', 'parent', 'show', 'url', 'priority', 'permission_id']

 admin.site.register(api_models.Menus, MenusAdmin)

#  模版中件件，用于在页面返回且将菜单渲染出来

def make_menus_html(menus, parent_id=None, current_parent_id=None, active=None):
    """
    menus = Menus.objects.all()
    :param menus: 寻找的对象，传一个queryset对象
    :param parent_id: 父级菜单ID
    :param current_parent_id: 当前父级菜单ID
    :param active: 激活的菜单名
    :return:
    """
    make_html = ""
    for menu in menus:
        child_menu_flag = "treeview"
        menu_right_flag = '<span class="pull-right-container"><i class="fa fa-angle-left pull-right"></i></span>'
        child_menu = '<li><a href="{menu_url}"><i class="fa fa-circle-o"></i> {menu_name}</a></li>'
        child_menu_html = '<ul class="treeview-menu">{make_child_menu_html}</ul>'
        master_menu_html = """
        <li class="{child_menu_flag} {active}">
            <a href="{menu_url}"><i class="fa {menu_icon}"></i> <span>{menu_name}</span>{menu_right_flag}</a>
            <ul class="treeview-menu">
            {children_menu_html}
            </ul>
        </li>"""
        children_menu_html = """
        <li class="treeview">
            <a href="{menu_url}"><i class="fa fa-circle-o"></i> <span>{menu_name}</span>{menu_right_flag}</a>
            {child_menu_html}
        </li>"""
        parent = menu.parent  # 获取当前菜单的父级菜单
        if current_parent_id == menu.id or (not parent and current_parent_id):
            continue  # 如果当前父级菜单ID是自己或没有父级菜单且有当前父级ID则跳过本次循环
        if not parent and current_parent_id is None:  # 如果没有父级菜单且当前父级ID是None
            make_children_menu_html = make_menus_html(menus, parent_id=parent_id, current_parent_id=menu.id)
            if not make_children_menu_html:
                menu_right_flag = ''
            menu_icon = "fa-eye"
            if hasattr(menu, 'icon_name'):
                menu_icon = menu.icon_name
            if menu.name == active:
                active_menu = 'active'
            else:
                active_menu = ''
            make_master_menu_html = master_menu_html.format(child_menu_flag=child_menu_flag,
                                                            active=active_menu,
                                                            menu_url=menu.url,
                                                            menu_icon=menu_icon,
                                                            menu_name=menu.name,
                                                            menu_right_flag=menu_right_flag,
                                                            children_menu_html=make_children_menu_html)
            make_html += make_master_menu_html
        elif parent and current_parent_id == parent.id:  # 如果有父级且当前父级ID是自己的父级ID
            make_child_menu_html = make_menus_html(menus, parent_id=current_parent_id, current_parent_id=menu.id)
            if make_child_menu_html:
                child_menu_html = child_menu_html.format(make_child_menu_html=make_child_menu_html)
                children_menu_html = children_menu_html.format(menu_url=menu.url,
                                                               menu_name=menu.name,
                                                               menu_right_flag=menu_right_flag,
                                                               child_menu_html=child_menu_html)
            else:
                children_menu_html = child_menu.format(menu_url=menu.url, menu_name=menu.name)
            make_html += children_menu_html
        else:
            continue
    return make_html

def make_menus_processor(request):
    menus_obj = Menus.objects.all()
    menus = make_menus_html(menus=menus_obj, active="监控")
    return {'menus': format_html(menus)}

然后通过Django  admin后台添加菜单数据，即可实现层叠动态可折叠菜单，该样式模版基于一个开源的管理模版

https://github.com/almasaeed2010/AdminLTE