Get single record from database Django - django

Question

I am new to Django and trying to make a basic website and having a few difficulties getting the relevant record for a page.
Models.py
import datetime
from django.db import models
from django.conf import settings
from ckeditor.fields import RichTextField
from markdown import markdown
class LiveEntryManager(models.Manager):
def get_query_set(self):
return super(LiveEntryManager,self).get_query_set().filter(status=self.model.LIVE_STATUS)
class Page(models.Model):
LIVE_STATUS = 1
HIDDEN_STATUS = 2
STATUS_CHOICES = (
(LIVE_STATUS, 'Live'),
(HIDDEN_STATUS, 'Hidden'),
)
title = models.CharField(max_length=250, help_text='Max 250 characters.')
slug = models.SlugField(unique=True, help_text='Suggested automatically generated from the title. Must be unique')
description = RichTextField()
description_html = models.TextField(editable=False, blank=True)
status = models.IntegerField(choices=STATUS_CHOICES, default=LIVE_STATUS,
help_text="Only pages with live status will be publicly displayed")
def save(self, force_insert=False, force_update=False):
self.description_html = markdown(self.description)
super(Page, self).save(force_insert, force_update)
def get_record(self):
return self.objects.get()
#Managers
live = LiveEntryManager()
objects = models.Manager()
class Meta:
ordering = ['title']
verbose_name_plural = "Pages"
def __unicode__(self):
return self.title
def get_absolute_url(self):
return "/%s/" % self.slug
class Image(models.Model):
page = models.ForeignKey(Page)
name = models.CharField(max_length=250)
image = models.ImageField(upload_to='gallery')
class Meta:
ordering = ['name']
def __unicode__(self):
return self.name
And Views.py
from django.shortcuts import get_object_or_404, render_to_response
from django.views.generic.list_detail import object_list
from mainsite.models import Page, Image
def home(request):
return render_to_response('templates/home.html')
def page(request, slug):
one_page = get_object_or_404(Page, slug=slug)
return render_to_response('templates/page/page.html',
{ 'object_list': one_page.get_record() })
and Urls.py
urlpatterns = patterns('',
# Examples:
url(r'^$', 'mainsite.views.home', name='home'),
# url(r'^disability/', include('disability.foo.urls')),
url(r'^page/(?P<slug>[-\w]+)/$', 'mainsite.views.page'),
# Uncomment the admin/doc line below to enable admin documentation:
# url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
url(r'^grappelli/', include('grappelli.urls')),
# Uncomment the next line to enable the admin:
url(r'^admin/', include(admin.site.urls)),
)
What I want to do is to be able to go to say /page/test and get the title, description and image for that object only. I know how to get all objects and display them in a template but cannot figure out how to get a single record and would be grateful of some help.
Thanks

Answer 1

If you would want to query the database for a single Page model you could use
views.py
from django.core.exceptions import ObjectDoesNotExist
from django.http import HttpResponse
from models import Pages
def page(request, slug):
try:
page = Pages.objects.get(slug=slug)
except ObjectDoesNotExist:
return HttpResponse(status=404)
return render(request, 'templates/page/page.html', {'page': page})
Then just use
{{page.title}}
or something like that in your template.
Also, if you want to query and have an iterable list returned intead you coule use
Pages.objects.filter()
instead of
Pages.objects.get()

Answer 2

Where'd you get get_record() from? The return value of get_object_or_404 is a model instance. There's nothing else to "get", just pass the instance to your context. And, don't use object_list as a context name. Not only is this an instance, not a list or queryset, but that name is most often used with pagination, and you'll simply create confusion in your code if you're not actually paginating anything.
return render_to_response('templates/page/page.html', {
'one_page': one_page,
})
Then in your template:
<h1>{{ one_page.title }}</h1>
<img src="{{ one_page.image.url }}" alt="">
<p>{{ one_page.description }}</p>