标签: render-to-response

我to_json在一个对象上使用该方法,并试图让:methods参数工作.我的模型(Drop)上有一个方法叫做is_favorited_by_user?.此方法接受一个参数current_user,然后检查用户是否收到了drop.如何通过该to_json方法传递此参数.

render :json => @drops.to_json(:methods => :is_favorited_by_user?(current_user))

嗨,我收到此错误,而我尝试制作一个简单的应用程序.

NameError at /first/

global name 'render_to_response' is not defined

Request Method:     GET
Request URL:    http://localhost:8000/first/?preview
Django Version:     1.3
Exception Type:     NameError
Exception Value:    

global name 'render_to_response' is not defined

Exception Location:     /home/naveen/django_projects/myproject/first/views.py in index, line 5
Python Executable:  /usr/bin/python
Python Version:     2.6.6
Python Path:    

['/home/naveen/django_projects/myproject',
 '/usr/local/lib/python2.6/dist-packages/pip-0.8.3-py2.6.egg',
 '/usr/local/lib/python2.6/dist-packages',
 '/usr/local/lib/python2.6/dist-packages/django_evolution-0.6.2-py2.6.egg',
 '/usr/lib/python2.6',
 '/usr/lib/python2.6/plat-linux2',
 '/usr/lib/python2.6/lib-tk',
 '/usr/lib/python2.6/lib-old',
 '/usr/lib/python2.6/lib-dynload',
 '/usr/local/lib/python2.6/dist-packages',
 '/usr/lib/python2.6/dist-packages',
 '/usr/lib/python2.6/dist-packages/PIL',
 '/usr/lib/python2.6/dist-packages/gst-0.10',
 '/usr/lib/pymodules/python2.6',
 '/usr/lib/python2.6/dist-packages/gtk-2.0',
 '/usr/lib/pymodules/python2.6/gtk-2.0']

Server time:    Thu, 31 Mar 2011 14:50:32 -0500

有任何想法吗？!

如何在Django render_to_response中使用HTTP响应头

我想使用Cache-Control,但无法判断它是否正常工作.以下是正确的:

render_to_response(templatename, {'Cache-Control':'no-cache'},context_instance=RequestContext(httpreq))

我正在为Django的render_to_response()编写一个包装函数来添加CSRF处理.

逻辑是:

def some_view (request)
    dictionary = {'context_param': some_param}
    dictionary.update(csrf(request))
    # ... view code here
    return render_to_response("a_template.html", dictionary)

render_to_response()具有以下签名:

def render_to_response(*args, **kwargs)

我想写透明包装器 - 只需添加一些功能(前面提到过)并保留其他内容.我想我应该写这样的东西:

def csrf_response (request, *args, **kwargs):

    # Here I need to somehow extract dictionary from args or kwargs

    if dictionary is None:
        dictionary = {}
    dictionary.update(csrf(request))

    # And here I somehow need to pass dictionary into render_to_response() fot the furher processing
    return render_to_response(*args, **kwargs)

所以问题是 - 从args/kwargs中提取所需参数(然后更改它)并将其进一步传递的最佳实践是什么？

render_to_response()的BTW代码似乎有点奇怪.这里是:

def render_to_response(*args, **kwargs):
    """
    Returns a HttpResponse whose …

您好我想要提供我内容的纯文本版本.所以我有一个单独的模板.我打电话render_to_response用mimetype="text/plain",但我想告诉浏览器中打开该页面的HTTP响应内容是UTF-8编码.我该怎么做(例如我需要添加什么render_to_response)？

您好,并提前感谢您.

我知道这是完全noob问题,我已经在各种论坛中搜索并阅读并重新阅读了文档,所以请保持温和.

我有一个观点:

#views.py

from django.shortcuts import render_to_response
from django.shortcuts import render
from django.http import HttpResponse, HttpRequest, HttpResponseRedirect
from acme.acmetest.models import Player
from acme.acmetest.models import PickForm

def playerAdd(request, id=None):
    form = PickForm(request.POST or None,
                       instance=id and Player.objects.get(id=id))

    # Save new/edited pick
    if request.method == 'POST' and form.is_valid():
        form.save()
        return HttpResponseRedirect('/draft/')

    #return render_to_response('makepick.html', {'form':form})
    return render(request, 'makepick.html', {'form':form})

def draftShow(request):
    draft_list = ['1', 'hello', 'test', 'foo', 'bar']
    #draft_list = Player.objects.all()
    #return render_to_response('makepick.html', {'draft_list' :draft_list}, context_instance=RequestContext(request))
    return render_to_response('makepick.html', {'draft_list' :draft_list})

我试图让它渲染到模板.html页面:

#makepick.html …

我正在Django中尝试一个简单的用户注册表单.我使用时没有生成CSRF令牌render_to_response

return render_to_response('registration/register.html',RequestContext(request,{'form':RegistrationForm()}))

在我使用的时候生成CSRF令牌 render

return render(request,'registration/register.html',{'form':RegistrationForm()})

我做错了什么 render_to_response

以下是相关的代码块

views.py

@csrf_protect
def register(request):
    if request.method == 'POST':
        form = RegistrationForm(request.POST)
        if form.is_valid():
            user = User.objects.create_user(
                username=form.cleaned_data['username'],
                password=form.cleaned_data['password1'],
                email=form.cleaned_data['email']
            )
            return HttpResponseRedirect('/register_success/')
    else:
        return render_to_response('registration/register.html', RequestContext(request, {'form': RegistrationForm()}))

register.html

{% extends "base.html" %}
{% block title %}User Registration{% endblock %}

{% block content %}
    <form method="post" action=".">
        {% csrf_token %}
        <table border="0">
            {{ form.as_table }}
        </table>
        <input type="submit" value="Register" />
    </form>
{% endblock %}

我有一个字符串,它将显示在html文件中.字符串中的某些单词(标记为"spc")需要以黄色背景和较大的字体显示.

我试图使用render_to_response方法将字符串(称为tdoc)发送到html文件.我用div标签替换了字符串中的'spc'标签.假设,在更换后,部分字符串是 we would seldom be prepared to <div id="spcl">examine</div> every.我的django代码看起来像render_to_response('a.html',{'taggeddoc':tdoc})

在我的CSS中,我有以下代码

 #spcl {  
background-color: #FFFF00;  
font-size:15px;  
}  

所以,我应该看到粗体字和黄色背景中的单词检查,但我没有看到.当我查看渲染的html的源代码时,它具有以下子字符串We would seldom be prepared to <div id="spcl">examine</div> every而不是原始字符串.

如何以所需方式显示"检查"和类似词语？

有哪些逻辑选择:

{% if request.user.is_active %}

如果视图没有render_to_response带字典返回request,则模板无法正常工作.

所以,欢迎任何想法.

//编辑

我的动机是:当您处理已安装应用程序的视图时,您没有机会修改它们.