Django動態渲染多層菜單

 

爲後續給菜單設置權限管理方便，經過給頁面模版菜單動態渲染，經過數據菜單表進行匹配須要渲染的菜單

 1 #Django表結構
 2 
 3 class Menus(models.Model):
 4 
 5     name = models.CharField(max_length=32, verbose_name=u'菜單名')
 6     parent = models.ForeignKey('self',
 7                                verbose_name=u'父級菜單',
 8                                null=True,
 9                                blank=True,
10                                default='0',
11                                help_text=u'若是添加的是子菜單，請選擇父菜單')
12     show = models.BooleanField(verbose_name=u'是否顯示',
13                                default=False,
14                                help_text=u'菜單是否顯示，默認添加不顯示')
15     url = models.CharField(max_length=300,
16                            verbose_name=u'菜單url地址',
17                            null=True,
18                            blank=True,
19                            default='javascript:void(0)',
20                            help_text=u'是否給菜單設置一個url地址')
21     priority = models.IntegerField(verbose_name=u'顯示優先級',
22                                    null=True,
23                                    blank=True,
24                                    default=-1,
25                                    help_text=u'菜單的顯示順序，優先級越大顯示越靠前')
26     permission_id = models.IntegerField(verbose_name=u'權限編號',
27                                         help_text=u'給菜單設置一個編號，用於權限控制',
28                                         error_messages={'field-permission_id': u'只能輸入數字'})
29 
30     def __str__(self):
31         return "{parent}{name}".format(name=self.name, parent="%s-->" % self.parent.name if self.parent else '')
32 
33     class Meta:
34         verbose_name = u"菜單"
35         verbose_name_plural = u"菜單"
36         ordering = ["-priority", "id"]

 1 #admin.py
 2 
 3 from django.contrib import admin
 4 
 5 # Register your models here.
 6 
 7 from cloud.api import models as api_models
 8 
 9 
10 class MenusAdmin(admin.ModelAdmin):
11     ordering = ('-parent',)
12     list_filter = ('name',)
13     list_display = ['name', 'parent', 'show', 'url', 'priority', 'permission_id']
14     fields = ['name', 'parent', 'show', 'url', 'priority', 'permission_id']
15 
16 admin.site.register(api_models.Menus, MenusAdmin)

#  模版中件件，用於在頁面返回且將菜單渲染出來

def make_menus_html(menus, parent_id=None, current_parent_id=None, active=None):
    """
    menus = Menus.objects.all()
    :param menus: 尋找的對象，傳一個queryset對象
    :param parent_id: 父級菜單ID
    :param current_parent_id: 當前父級菜單ID
    :param active: 激活的菜單名
    :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  # 獲取當前菜單的父級菜單
        if current_parent_id == menu.id or (not parent and current_parent_id):
            continue  # 若是當前父級菜單ID是本身或沒有父級菜單且有當前父級ID則跳過本次循環
        if not parent and current_parent_id is None:  # 若是沒有父級菜單且當前父級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:  # 若是有父級且當前父級ID是本身的父級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="監控")
    return {'menus': format_html(menus)}

 

 

而後經過Django  admin後臺添加菜單數據，便可實現層疊動態可摺疊菜單，該樣式模版基於一個開源的管理模版

https://github.com/almasaeed2010/AdminLTE