Django--admin源码流程

原创

qq62009e0d0a7d2 2022-03-31 16:23:20 博主文章分类：Django ©著作权

©著作权归作者所有：来自51CTO博客作者qq62009e0d0a7d2的原创作品，请联系作者获取转载授权，否则将追究法律责任

admin.py

from django.contrib import admin
from . import models
"""
通过原生的django admin来创造数据
"""
admin.site.register(models.User)
admin.site.register(models.Role)
admin.site.register(models.Permission)
admin.site.register(models.Group)
admin.site.register(models.Menu)

url.py

from django.conf.urls import url
from django.contrib import admin
from app01 import views
urlpatterns = [
    url(r'^admin/', admin.site.urls),
]

以上两个文件都引用的是django.contrib.admin.site中的方法，site是AdminSite实例化出来的一个对象，所以site调用的register方法和urls属性(方法被@property装饰器装饰)是AdminSite这个类中定义的

django启动的时候，顺序为：先register，然后在urls分配

先看admin.site.register干了什么

在源码AdminSite这个类的构造方法初始化定义了一个空字典 _registry={}

12345678	`classAdminSite(object):` `...` `def__init__(self, name='admin'):` `self._registry` `={}` `# model_class class -> admin_class instance` `self.name` `=name` `self._actions` `={'delete_selected': actions.delete_selected}` `self._global_actions` `=self._actions.copy()` `all_sites.add(self)`

看源码中的admin.site.register函数如下

def register(self, model_or_iterable, admin_class=None, **options):
        """
        Registers the given model(s) with the given admin class.

        The model(s) should be Model classes, not instances.

        If an admin class isn't given, it will use ModelAdmin (the default
        admin options). If keyword arguments are given -- e.g., list_display --
        they'll be applied as options to the admin class.

        If a model is already registered, this will raise AlreadyRegistered.

        If a model is abstract, this will raise ImproperlyConfigured.
        """
        if not admin_class:
            admin_class = ModelAdmin

        if isinstance(model_or_iterable, ModelBase):
            model_or_iterable = [model_or_iterable]
        for model in model_or_iterable:
            if model._meta.abstract:
                raise ImproperlyConfigured(
                    'The model %s is abstract, so it cannot be registered with admin.' % model.__name__
                )

            if model in self._registry:
                raise AlreadyRegistered('The model %s is already registered' % model.__name__)

            # Ignore the registration if the model has been
            # swapped out.
            if not model._meta.swapped:
                # If we got **options then dynamically construct a subclass of
                # admin_class with those **options.
                if options:
                    # For reasons I don't quite understand, without a __module__
                    # the created class appears to "live" in the wrong place,
                    # which causes issues later on.
                    options['__module__'] = __name__
                    admin_class = type("%sAdmin" % model.__name__, (admin_class,), options)

                # Instantiate the admin class to save in the registry
                self._registry[model] = admin_class(model, self)

register函数第一个参数是注册的模块名称，第二个参数不传默认是None，但是实际使用的是ModelAdmin，ModelAdmin已模块名称作为参数实例化得到的对象作为 registry字典的value。简化如下：

def register(self, model_or_iterable, admin_class=None, **options):
    if not admin_class:
        admin_class = ModelAdmin

    for model in model_or_iterable:
        .......
        .......
        self._registry[model] = admin_class(model, self)

再看admin.site.urls干了什么

源码精简如下：

def get_urls(self):
    from django.conf.urls import url, include
    from django.contrib.contenttypes import views as contenttype_views

    def wrap(view, cacheable=False):
        ......
        return update_wrapper(wrapper, view)

    # Admin-site-wide views.
    urlpatterns = [
        url(r'^$', wrap(self.index), name='index'),
        url(r'^login/$', self.login, name='login'),
        url(r'^logout/$', wrap(self.logout), name='logout'),
        url(r'^password_change/$', wrap(self.password_change, cacheable=True), name='password_change'),
        ......
    ]

    valid_app_labels = []
    for model, model_admin in self._registry.items():
        urlpatterns += [
            url(r'^%s/%s/' % (model._meta.app_label, model._meta.model_name), include(model_admin.urls)),
        ]
        if model._meta.app_label not in valid_app_labels:
            valid_app_labels.append(model._meta.app_label)
        ......
        
    return urlpatterns

@property
def urls(self):
    return self.get_urls(), 'admin', self.name #返回的是一个元组

首先urls是个函数被@property装饰器装饰为属性，urls属性返回的是一个元组【元组的第一个元素是列表，由get_urls函数返回】

get_urls函数中定义了一个urlpatterns并最终返回，里面有url路径和视图函数的对应关系，这些默认register(admin)后自带的，所以在urls.py 执行

1	`url(r'^admin/', admin.site.urls)`

会找到注册的admin类，为每一个类生成N个URL（如上urlpattern中定义的）

如果用户自己注册的模块，如下

1	`admin.site.register(models.UserInfo)`　`# 注册的时候没有传入admin_class,默认使用的是ModelAdmin`

会根据下面的代码添加到urlpatterns中

for model, model_admin in self._registry.items():
    urlpatterns += [
        url(r'^%s/%s/' % (model._meta.app_label, model._meta.model_name), include(model_admin.urls)),
    ]

model_class._meta.app_label 代表model所在的模块名比如app01

model_class._meta.model_name 代表model对应的表的小写名，比如userinfo

这样就生成了/app01/userinfo/ 这样的url前缀。

后面调用的include(model_admins.urls) 中的model_admins.urls 即执行的是self._registry['UserInfo'].urls ===== admin_class(model, self).urls=======admin_class这个类生成对象的urls属性【此时这个对象是UserInfo的对象，所以下面get_url函数中的self指的就是UserInfo的对象】======admin_class这个类的urls属性========= > ModelAdmin这个类的urls属性【因为注册的时候没有传入admin_class，所以默认是ModelAdmin】

再来看下ModelAdmin这个类的urls属性

def get_urls(self):
    from django.conf.urls import url

    def wrap(view):
        def wrapper(*args, **kwargs):
            return self.admin_site.admin_view(view)(*args, **kwargs)
        wrapper.model_admin = self
        return update_wrapper(wrapper, view)

    info = self.model._meta.app_label, self.model._meta.model_name

    urlpatterns = [
        url(r'^$', wrap(self.changelist_view), name='%s_%s_changelist' % info),
        url(r'^add/$', wrap(self.add_view), name='%s_%s_add' % info),     # self代表的是UserInfo的对象
        url(r'^(.+)/history/$', wrap(self.history_view), name='%s_%s_history' % info),
        url(r'^(.+)/delete/$', wrap(self.delete_view), name='%s_%s_delete' % info),
        url(r'^(.+)/change/$', wrap(self.change_view), name='%s_%s_change' % info),
        # For backwards compatibility (was the change url before 1.9)
        url(r'^(.+)/$', wrap(RedirectView.as_view(
            pattern_name='%s:%s_%s_change' % ((self.admin_site.name,) + info)
        ))),
    ]
    return urlpatterns

@property
def urls(self):
    return self.get_urls()

同理，urls是个被@property装饰的属性，调用ModelAdmin类自身的get_urls函数的时候返回的即是上面红色的urlpatterns【这也是django admin为什么注册了一个类之后就为这个类提供了基本url和函数映射的基本原因】，这样和上面形成的url路径拼接即可得到如下的效果

Django--admin源码流程_元组

urlpatterns = [
        url(r'admin/app01/userinfo/', wrap(self.changelist_view), name='%s_%s_changelist' % info),
        url(r'admin/app01/userinfo/add/$', wrap(self.add_view), name='%s_%s_add' % info),        # self 代表的是UserInfo
        url(r'admin/app01/userinfo/(.+)/history/$', wrap(self.history_view), name='%s_%s_history' % info),
        url(r'admin/app01/userinfo/(.+)/delete/$', wrap(self.delete_view), name='%s_%s_delete' % info),
        url(r'admin/app01/userinfo/(.+)/change/$', wrap(self.change_view), name='%s_%s_change' % info),
        # For backwards compatibility (was the change url before 1.9)
        url(r'admin/app01/userinfo/(.+)/$', wrap(RedirectView.as_view(
            pattern_name='%s:%s_%s_change' % ((self.admin_site.name,) + info)
        ))),
    ]

Django--admin源码流程_perl_02

当然也可以自定制ModelAdmin，如下

from django.contrib import admin
from django.contrib.admin import ModelAdmin
from . import models
from django.shortcuts import HttpResponse

class UserInfoModelAdmin(ModelAdmin):

# 没有__super__ 是这个类只定义了一个方法，没有继承其他的？还是继承了所有，只覆盖了这一个呢？

# 自定义功能
    def changelist_view(self, request, *args,**kwargs):
        return HttpResponse('用户列表')

admin.site.register(models.UserInfo,UserInfoModelAdmin)

class UserTypeModelAdmin(ModelAdmin):
    pass
    
admin.site.register(models.UserType,UserTypeModelAdmin)

继承自ModelAdmin，并重写了changelist_view 这个方法，父类其他的继承了吗？==

所以上面这张register的方法可以如下理解：

在admin.site对象的 _registry = {}
_registry = {
    models.UserInfo: obj1 = ModelAdmin(models.UserInfo,admin.site),   #obj1 即上面说的UserInfo对象
    /admin/app01/userinfo/                obj1.changelist_view
    /admin/app01/userinfo/add/            obj1.add_view
    /admin/app01/userinfo/(\d+)/delete/   obj1.delete_view
    /admin/app01/userinfo/(\d+)/change/   obj1.change_view
                                          self.model=models.UserInfo


    models.UserType: obj2 = ModelAdmin(models.UserType,admin.site),  #obj2 和上面说的同理，代表的是UserType对象
    /admin/app01/usertype/                obj2.changelist_view
    /admin/app01/usertype/add/            obj2.add_view
    /admin/app01/usertype/(\d+)/delete/   obj2.delete_view
    /admin/app01/usertype/(\d+)/change/   obj2.change_view
                                          self.model=models.UserType
}

for model, model_admin in self._registry.items():
    urlpatterns += [
        # url(r'^%s/%s/' % (model._meta.app_label, model._meta.model_name), include(model_admin.urls)),
        url(r'^%s/%s/' % ('app01', 'userinfo'), include(model_admin.urls)),
    ]