小编cho*_*obo的帖子

我想request.META.get('HTTP_REFERER')在模板中使用.

我的模板来源:

<!-- this is login.html -->
{% extends "base.html" %}
{% block title %}django bookmark- login{% endblock %}
{% block head %}login{% endblock %}
{% block content %}
    {% if form.errors %}
    <p>try again!</p>
    {% endif %}
    <form method="post" action=".">{% csrf_token %}
        <p><label for="id_username">username:</label>
        {{ form.username }}</p>
        <p><label for="id_password">password:</label>
        {{ form.password }}</p>
        <input type="hidden" name="next" value="/<!-- I WANT TO USE 'HTTP_REFERER' HERE -->" />
        <input type="submit" value="login" />
    </form>
{% endblock %}

我该怎么办？

urlpatterns = patterns('', (r'^login/$', …

如何将多行连接成一个字符串？

查询:

SELECT name FROM mytable;

结果:

name
----
kim
lee
park
cho

只是我想要.

name
----
kim,lee,park,cho

不可能？

如何在django-rest-framework的modelviewset中取消设置csrf？

我将使用django-rest-framework的viewsets.ModelViewSet(http://django-rest-framework.org/api-guide/viewsets.html#modelviewset).

我的应用程序是api服务器.所以我不需要使用csrf.

但我不知道如何取消csrf.

请举个例子!

https://gist.github.com/meooo​​h/896b66e0fc5c9bfd5716fc1864a7c68b

如果我有两个 iam 角色。

但我不知道区别

我想在views.py中调用基于类的通用视图

请看我的代码......

urls.py

from django.conf.urls import patterns, include, url
from crm.views import *

urlpatterns = patterns('',
    (r'^workDailyRecord/$', workDailyRecord),
)

和我的views.py ....请看...

views.py

from django.views.generic import TodayArchiveView
from crm.models import *

def workDailyRecord(request):
    if request.method == 'GET':
        tView.as_view() # I want call class-based generic views at this line.
    elif:
        """
        Probably this part will be code that save the data.
        """
        pass

class tView(TodayArchiveView):
    model = WorkDailyRecord
    context_object_name = 'workDailyRecord'
    date_field = 'date'
    template_name = "workDailyRecord.html"

我该怎么办？

Django1.4:如何在模板中使用order_by？

models.py

from django.db import models
from django.contrib.auth.models import User
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic

class Note(models.Model):
    contents = models.TextField()
    writer = models.ForeignKey(User, to_field='username')
    date = models.DateTimeField(auto_now_add=True)

    content_type = models.ForeignKey(ContentType)
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey('content_type', 'object_id')


class Customer(models.Model):
    name = models.CharField(max_length=50,)
    notes = generic.GenericRelation(Note, null=True)

以上是我的models.py.

我想使用'order_by'(https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by)

和...

views.py

from django.views.generic import DetailView
from crm.models import *

class customerDetailView(DetailView):
    context_object_name = 'customerDetail'
    template_name = "customerDetail.html"
    allow_empty = True
    model = Customer
    slug_field = 'name'

我的views.py使用DetailView( …

我看完了文档的reverse()Django的URL调度的方法.

什么时候有用？

谢谢!

我想将一个数字传递给我的通用视图(DetailView)来获取一个对象这是我的代码

URLPATTERN

(r'^newreportview/(?P<number>\w+)/$', NewReportView.as_view()),  

查看课程

class NewReportView(DetailView):
    template_name = "report/newreportview.html"
    context_object_name = "newreportview"
    def get_queryset(self):
        task= get_object_or_404(MyTask,applicationnnumber=self.args[0])
        return task

我猜这一行有些不对劲

name = get_object_or_404(MyTask,applicationnnumber=self.args[0])

错误信息:

Exception Type: IndexError
Exception Value:
tuple index out of range

我应该如何将'number'传递给这个通用视图并获取带有这个'number'的Mytask对象？

谢谢