Question

当我在django admin中保存模型时，我需要一种显示中间页面的方法。

我想要完成的是在“保存”模型之后，显示一个页面，其中列出了模型的所有属性，然后有一个显示Print的按钮。单击save时，我曾经使用Jquery对话框div来解决这个问题。这意味着我在实际保存模型之前显示了设置打印视图，但我现在需要模型首先验证。

它就像“删除模型”操作的实现方式一样。我似乎无法找到从哪里开始寻找。

修改我开始在django.contrib.admin.options.py查找response_change和response_add方法。不知道如何覆盖它们。它只需要一个特定的模型，因此它不是通用的。我还发现了类ModelAdmin中的模板列表。仍然不确定如何在不破坏管理员的情况下继续进行。

编辑2： 在下面添加了我的工作解决方案。

Answer 1

您可以创建一个表单，对“您确定”步骤进行额外的验证步骤。

在models.py中给出了这个模型：

from django.db import models

class Person(models.Model):
    name = models.CharField(max_length=100)

在forms.py中添加表单：

from django import forms
from .models import Person

class PersonForm(forms.ModelForm):
    i_am_sure = forms.BooleanField(required=False, widget=forms.HiddenInput())

    def __init__(self, *args, **kwargs):
        super(PersonForm, self).__init__(*args, **kwargs)

        if self.errors.get('i_am_sure'):
            # show the 'are you sure' checkbox when we want confirmation
            self.fields['i_am_sure'].widget = forms.CheckboxInput()

    def clean(self):
        cleaned_data = super(PersonForm, self).clean()

        if not self.errors:
            # only validate i_am_sure once all other validation has passed
            i_am_sure = cleaned_data.get('i_am_sure')
            if self.instance.id and not i_am_sure:
                self._errors['i_am_sure'] = self.error_class(["Are you sure you want to change this person?"])
                del cleaned_data['i_am_sure']

        return cleaned_data

    class Meta:
        model = Person

如果你想在Django admin中使用它。在admin.py中指定此表单：

from django.contrib import admin
from .forms import PersonForm
from .models import Person

class PersonAdmin(admin.ModelAdmin):
    form = PersonForm

admin.site.register(Person, PersonAdmin)

但请注意，Django管理表单上存在隐藏输入的错误。 There's a solution to that on this Stack Overflow question

Answer 2

您可以向ModelAdmin添加视图和网址，并覆盖您的modeladmin添加视图以进行相应的重定向。

class MyModelAdmin(admin.ModelAdmin):
    def get_urls(self):
        urls = super(MyModelAdmin, self).get_urls()
        my_urls = patterns('',
            (r'^my_view/$', self.my_view)
        )
        return my_urls + urls

    def my_view(self, request):
        # custom view which should return an HttpResponse
        pass

Answer 3

所以，经过一些编码后我就开始工作了。

我的modeladmin现在看起来像这样

class MyModelAdmin(admin.ModelAdmin):
    def get_urls(self):
        urls = super(MyModelAdmin, self).get_urls()
        my_urls = patterns('',
            (r'^my_view/$', self.my_view)
        )
        return my_urls + urls

    def my_view(self,request,pk):
        from django.shortcuts import render_to_response
        from django.template import RequestContext

        object = Model.objects.get(pk=pk)
        model_dict = model_object.__dict__
        return render_to_response('admin/app_name/model/model_view.html',locals(),context_instance=RequestContext(request))

    @csrf_protect_m
    @transaction.commit_on_success
    def add_view(self, request, form_url='', extra_context=None):
        "The 'add' admin view for this model."
        model = self.model
        opts = model._meta

        if not self.has_add_permission(request):
            raise PermissionDenied

        ModelForm = self.get_form(request)
        formsets = []
        inline_instances = self.get_inline_instances(request)
        if request.method == 'POST':
            form = ModelForm(request.POST, request.FILES)
            if form.is_valid():
               new_object = self.save_form(request, form, change=False)
               form_validated = True
            else:
               form_validated = False
               new_object = self.model()
            prefixes = {}
            for FormSet, inline in zip(self.get_formsets(request), inline_instances):
               prefix = FormSet.get_default_prefix()
               prefixes[prefix] = prefixes.get(prefix, 0) + 1
               if prefixes[prefix] != 1 or not prefix:
                  prefix = "%s-%s" % (prefix, prefixes[prefix])
               formset = FormSet(data=request.POST, files=request.FILES,
                              instance=new_object,
                              save_as_new="_saveasnew" in request.POST,
                              prefix=prefix, queryset=inline.queryset(request))
               formsets.append(formset)
            if all_valid(formsets) and form_validated:
               self.save_model(request, new_object, form, False)
               self.save_related(request, form, formsets, False)
               self.log_addition(request, new_object)
               log.info('The new object has %s id' % new_object.id)
               return HttpResponseRedirect('/admin/draws/contest/contest_view/%s' % new_object.id)  <-- changed to my new one
               .................
               .................

在templates/admin/app_name/model_view.html中创建了一个html模板，就是这样！

创建Django Admin Intermediate页面

3 个答案: