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