Django Admin-禁用编辑并删除特定型号的"保存"按钮

Question

我有一个Django模型,我希望它只是readonly.不允许添加和编辑.

我在readAdly中标记了所有字段,并在ModelAdmin中覆盖了has_add_permission:

class SomeModelAdmin(admin.ModelAdmin):
     def has_add_permission(self, request):
        return False

有类似的has_edit_permission吗？可以禁用哪个删除"保存"和"保存并继续"按钮？并通过简单的"关闭并返回"按钮替换.

Django文档仅提及只读字段而不是覆盖编辑权限.

Answer 1

对于Django 1.11:

def has_add_permission(self, request, obj=None):
    return False

def changeform_view(self, request, object_id=None, form_url='', extra_context=None):
    extra_context = extra_context or {}
    extra_context['show_save_and_continue'] = False
    extra_context['show_save'] = False
    return super(YourModelAdmin, self).changeform_view(request, object_id, extra_context=extra_context)

Answer 2

覆盖templates/admin/submit_line.html模板并根据需要制作按钮.您可以通过将其放入特定模型来执行此操作templates/admin/[app_label]/[model]/submit_line.html.

要有条件地显示默认提交行或您的自定义提交行,请覆盖ModelAdmin.change_view并添加一个布尔值extra_context:

class MyModelAdmin(admin.ModelAdmin):
    ...
    def change_view(self, request, object_id, extra_context=None):
        if not request.user.is_superuser:
            extra_context = extra_context or {}
            extra_context['readonly'] = True

        return super(MyModelAdmin, self).change_view(request, object_id, extra_context=extra_context)

Answer 3

我有同样的问题.我在admin.py中修复了它

from django.contrib.admin.templatetags.admin_modify import register, submit_row as original_submit_row

@register.inclusion_tag('admin/submit_line.html', takes_context=True)
def submit_row(context):
''' submit buttons context change '''
ctx = original_submit_row(context)
ctx.update({
    'show_save_and_add_another': context.get('show_save_and_add_another',
                                             ctx['show_save_and_add_another']),
    'show_save_and_continue': context.get('show_save_and_continue',
                                          ctx['show_save_and_continue']),
    'show_save': context.get('show_save',
                             ctx['show_save']),
    'show_delete_link': context.get('show_delete_link', ctx['show_delete_link'])
})
return ctx

在MyModelAdmin类中,添加以下函数

@classmethod
def has_add_permission(cls, request):
    ''' remove add and save and add another button '''
    return False

def change_view(self, request, object_id, extra_context=None):
    ''' customize add/edit form '''
    extra_context = extra_context or {}
    extra_context['show_save_and_continue'] = False
    extra_context['show_save'] = False
    return super(MyModelAdmin, self).change_view(request, object_id, extra_context=extra_context)

Answer 4

最简单的方法是禁用 ModelAdmin 类中的相应权限。例如，我有一个 Cart 模型，我希望管理员仅查看（列表和详细信息），而我所做的只是将以下函数添加到 CartAdmin 类以禁用删除、更改和添加

class CartAdmin(admin.ModelAdmin):
    list_display = ['listing']

    def has_add_permission(self, request, obj=None):
        return False

    def has_change_permission(self, request, obj=None):
        return False

    def has_delete_permission(self, request, obj=None):
        return False

has_add_permission、has_change_permission 和 has_delete_permission 三个方法是在admin 中禁用添加按钮、添加表单、编辑表单和删除按钮的方法

这是在禁用上述权限的管理员中查看记录详细信息时的示例输出

如上图所示，您只有关闭按钮，并且详细信息没有显示在表单中

Answer 5

使用Django 1.8(Python 3语法)更新了答案.

有三件事要做:
1)扩展管理员更改表单模板,添加if有条件地抑制提交按钮
2)覆盖admin.ModelAdmin.change_view()并设置模板的上下文var if以读取
3)禁止不需要的POST请求(来自DOM hacking,curl/Postman )

MyProject的/ my_app应用/模板/管理/程序my_app/change_form.html

{% extends "admin/change_form.html" %}
{% load admin_modify %}
{% block submit_buttons_top %}{% if my_editable %}{% submit_row %}{% endif %}{% endblock %}
{% block submit_buttons_bottom %}{% if my_editable %}{% submit_row %}{% endif %}{% endblock %}

MyProject/my_app/admin.py(MyModelAdmin)

def change_view(self, request, object_id, form_url='', extra_context=None):
  obj = MyModel.objects.get(pk=object_id)
  editable = obj.get_status() == 'Active'

  if not editable and request.method == 'POST':
    return HttpResponseForbidden("Cannot change an inactive MyModel")

  more_context = {
    # set a context var telling our customized template to suppress the Save button group
    'my_editable': editable,
  }
  more_context.update(extra_context or {})
  return super().change_view(request, object_id, form_url, more_context)

Answer 6

我有同样的问题 - 最简单的方法是包含一些自定义JS.

在你admin.py文件包含

class Media:
    js = ('/static/js/admin.js',)

然后在您的admin.js文件中,包含以下JS.

(function($) {
    $(document).ready(function($) {
         $(".submit-row").hide()
    });
})(django.jQuery);

这行已经消失了 - 它也适用于所有版本的Django.