I'm new at django and I'm having problems with a simple form POST.I have a ModelForm in forms.py and when user enters information in html, views.py taks it and saves it. However, I keep getting an error saying it can't find the view doesn't exist in view.py. Please help me find the error. Thank you!
urls.py
urlpatterns = patterns('',
(r'^mypage/(?P<username>\w+)/$', 'recipeapp.views.my_view'),
forms.py
class NewRecipeForm(forms.ModelForm):
user_info = forms.ForeignKey(User)
title = forms.CharField(min_length=2,max_length=50,required=True,)
post_date = forms.DateField(auto_now=True)
ingredients = forms.TextField(widget=forms.Textarea(),)
picture = forms.ImageField(upload_to='photos/%Y/%m/%d',)
content = forms.TextField(widget=forms.Textarea(),)
views.py
#csrf_protect
from recipeapp.forms import NewRecipeForm
def my_view(request,username):
if request.method == 'POST':
form = NewRecipeForm(request.POST)
if form.is_valid():
form.save()
else:
form = NewRecipeForm()
return render_to_response('postlogin.html',{'username':username},{'form': form}, RequestContext(request))
postlogin.html
<form action="" method="post" id="form">
{% csrf_token %}
<div id="dish-name">
<label><p>Dish name</p></label>
{{form.title}}
</div>
<div id="ingredients">
<label><p>Ingredients</p></label>
{{form.ingredients}}
</div>
<div id="content">
<label><p>Content</p></label>
{{form.content}}
</div>
{{form.picture}}
</form>
Is that really your whole views.py? You have at least three issues:
Firstly, you haven't imported csrf_protect - like any name, a decorator needs to be defined before you can use it.
Secondly, you have to decorate an actual function, not a file. The decorator should go just before the function definition for my_view.
Thirdly, your indentation is broken - the def should not be indented at all.
Given all those, I expect that Python is failing to import your views because of syntax errors.
Also note that you shouldn't really use csrf_protect - you should enable CSRF protection in your middleware (it's on by default) and only use the csrf_exempt decorator, and then only on very very rare occasions.
Related
I'm making a simple web server to take html input data and insert it to data base table .
I've tried with POST request got into more CSRF troubles , Turned to GET request to go over CSRF (not necessary in my case ) , still not able to GET the html data into Database .
myapp/models.py
from django.db import models
class Post(models.Model):
title = models.CharField(max_length=300, unique=True)
content = models.TextField()
myapp/templates/createpost.html
<head>
<title>Create a Post </title>
</head>
<body>
<h1>Create a Post </h1>
<form action="" method="GET">
{%csrf_token%}
Title: <input type="text" name="title"/><br/>
Content: <br/>
<textarea cols="35" rows="8" name="content">
</textarea><br/>
<input type="submit" value="Post"/>
</form>
</body>
</html>
myapp/views.py
from django.shortcuts import render
from .models import Post
def createpost(request):
if request.method == 'GET':
if request.GET.get('title', None) and request.GET.get('content', None):
post = Post()
post.title = request.GET.get('title', None)
post.content = request.GET.get('content', None)
post.save()
return render(request, 'createpost.html')
else:
return render(request, 'createpost.html')
urls.py
from django.contrib import admin
from django.urls import path, include
from django.views.generic.base import TemplateView
urlpatterns = [
path('admin/', admin.site.urls),
path('', TemplateView.as_view(template_name='createpost.html'), name='createpost'),
]
am using Django 2.2.6 with PyCharm community 2019.2.3
I've been searching for almost 2 days , checked django doc , and stackoverflow answers , none was helpful , I had to ask to make sure its not a version related issue and forgive me if i miss understand a simple point am just a beginner.
I've tried with POST request got into more CSRF troubles , Turned to GET request to go over CSRF (not necessary in my case ) , still not able to make it .
Ok.
STOP EVERYTHING RIGHT NOW
Now
do the official Django tutorial and learn to use Django forms and modelforms
read the HTTP spec about the proper use of GET requests (hint: the important word here is "idempotent")
fix your code accordingly
if by then you still have issues with the csrf token, come back and post a question about it (with a proper MCVE etc).
Also note that your question should be claused as either unclear or OT since you didn't explain what your problem is (hint: "doesn't work" doesn't explain anything). But anyway...
What you want (nb: models unchanged) is:
myapp/forms.py
from django import forms
from . models import Post
class PostForm(forms.ModelForm):
class Meta(object):
model = Post
fields = ("title", "content")
views.py
from django.shortcuts import redirect, render
from . models import Post
from . forms import PostForm
def createpost(request):
if request.method == 'POST':
form = PostForm(request.POST)
if form.is_valid():
form.save()
# read the doc for `redirect` and change the destination to
# something that makes sense for your app.
# as to why we redirect, cf https://en.wikipedia.org/wiki/Post/Redirect/Get
return redirect("/")
else:
# GET request, present an empty form
form = PostForm()
return render(request, 'createpost.html', {"form": form})
urls.py
from django.contrib import admin
from django.urls import path, include
# NB: actually you should define myapp's urls in
# myapp/urls.py and `include` them here
from myapp.views import createpost
urlpatterns = [
path('admin/', admin.site.urls),
path('/', createpost, name='createpost'),
]
myapp/templates/createpost.html
<head>
<title>Create a Post </title>
</head>
<body>
<h1>Create a Post </h1>
<form action="" method="POST">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Post"/>
</form>
</body>
</html>
I can't garantee this will work out of the box (I'd have to post a full project with proper settings etc to be sure - and actually to test it at least once for typos etc) but that's the correct way of doing things.
I've been searching for almost 2 days
That's 2 days wasted, that you would have better used doing the official Django tutorial. I also suggest you learn to debug programs - just adding a couple ̀€print("XXX")` calls in your view and checking your devserver's outputs would have made clear why nothing got created in your database (hint: with your original urls.py, you are NEVER calling your view function).
I am using Django for develop a website. The website is intended to use to search information stored in a MySQL database.
This is the current basic flow of the web site.
1) index.html - this has a form to select an option
2) according the option, users will redirect to search.html (include a form)
3) once the user provides the criteria, the result will be displayed in reply.html
In my views.py , I have two functions.
from django.shortcuts import render
from website.models import WebsiteRepository
from .forms import SearchForm
from .forms import SelectTypeForm
def Search(request):
if request.method == 'POST':
#do something
return render(request, 'reply.html', {'env_dict':env_dict})
else:
#do something
return render(request, 'search.html', context = context)
def index(request):
if request.method =='POST':
#do something
return render(request, 'search.html', context = context)
else:
#do something
return render(request, 'index.html', context= context)
When I go to index.html page, I can select a option and it will direct me to search.html. After, I fill the form there and submit, it wont give me the reply.html page.
I have a feeling that, I could make this work by changing urls.py.
from django.urls import path
from website import views
urlpatterns = [
path('', views.index, name='index'),
#path('search/', view.Search, name ='Search')
]
I tried to google it. But its too much details and Iam kind of lost.
Do any of you guys know how to achieve this?
Thanks
search.html
{% extends "base_generic.html" %}
{% block content %}
<h3>Welcome to search information Repository</h3>
<form method="post">
{% csrf_token %}
{{form.as_p}}
<button type = 'submit'>submit</button>
</form>
{% endblock %}
index.html
{% block content %}
<h3>Welcome to information Repository</h3>
<form method="post">
{% csrf_token %}
{{form.as_p}}
<button type = 'submit'>submit</button>
</form>
just for clarify things more, ill add the forms.py too
from django import forms
from .models import WebsiteRepository
class SearchForm(forms.Form):
websiterepository = WebsiteRepository
env_indicators = websiterepository.objects.filter (key_aspect='Environmental').values_list('repo_id','indicator')
indicator = forms.ChoiceField(choices=env_indicators,label = 'Indicator' )
OPTIONS = (('2000','2000'),('2001','2001'),('2002','2002'), ('2003','2003'),('0000','0000'),)
year = forms.ChoiceField(choices=OPTIONS)
class SelectTypeForm(forms.Form):
OPTIONS = (('1', 'Envirnmental Indicators'),('2','Economic Indicators'),('3','Social Indicators'),)
types = forms.ChoiceField(choices=OPTIONS)
Your code is wrong on many points.
First thing first: for a search, you want a GET request, not a POST (POST is for updating the server's state - adding or updating your database mostly). This is the semantically correct method (since you want to GET data), and it will allow a user to bookmark the url.
Second point: you don't want to submit the search form to the index view but to the search view. No need for redirects etc, just use the {% url %} templatetag to fill the action attribute of your form (you of course need to have a 'Search' url in your urls.py):
<form method="get" action="{% url 'Search' %}">
{% csrf_token %}
{{form.as_p}}
<button type = 'submit'>submit</button>
</form>
if you want to have this form on more than one page (which is often the case for search forms), use an inclusion tag tha will take care of creating an unbound SearchForm and render the template fragment.
Then in your search view, you only want GET requests, and do not use two different templates, this will only lead to useless duplication.
def Search(request):
form = SearchForm(request.GET)
# use the form's data - if any - to get search results
# and put those results (even if empty) in you context
return render(request, 'reply.html', {'env_dict':env_dict})
And finally, your search form is totally broken:
class SearchForm(forms.Form):
# this is totally useless
websiterepository = WebsiteRepository
# this will only be evaluated once at process startup, so you will
# get stale data in production - and probably different data
# per process, in a totally unpredictable way.
# You need to either put this in the form's __init__ or wrap it
# in a callable and pass this callable
env_indicators = websiterepository.objects.filter (key_aspect='Environmental').values_list('repo_id','indicator')
indicator = forms.ChoiceField(choices=env_indicators,label = 'Indicator' )
# are you going to manually add a new year choice every year ???
OPTIONS = (('2000','2000'),('2001','2001'),('2002','2002'), ('2003','2003'),('0000','0000'),)
year = forms.ChoiceField(choices=OPTIONS)
For the "indicators" ChoiceField you want something like:
def get_indicators_choices():
return Websiterepository.objects.filter (key_aspect='Environmental').values_list('repo_id','indicator')
class SearchForm(forms.Form):
# IMPORTANT : we are NOT calling the function here, just
# passing it (python functions are objects) to the field, which
# will call it everytime the form is instanciated, so you don't
# have stale data
indicator = forms.ChoiceField(
choices=get_indicator_choices,
label='Indicator')
As a last note: be consistent with your namings (ie why name one view in all lower (index) and capitalize the other (Search) ? Whichever convention you choose (I strongly suggest respecting pep8 here), at least stick to it for the whole project.
The problem is that code is not redirecting to /search, instead rendering search.html after post from index.html.
Try doing like-
views.py-
#your code
def index(request):
#do something
if request.method == 'POST':
return redirect('Search')
else:
#render index.html
def search(request):
#do something
if request.method == 'POST':
#render reply.html
else:
#render search.html
Another way to achieve this is if you specify action in your form so that form posts on /search.
search.html
<form method="post" action="/search">
{% csrf_token %}
{{form.as_p}}
<button type = 'submit'>submit</button>
</form>
I'm trying to get the standard Aldryn Newsblog Buttons working in my Frontend Page. So every User can Add, Delete and Edit Articles(only the articles they created themselves but thats not the question). This is the Menu with the links:
Menu in the Toolbar
So i want to add a Button in my template wich triggers the edit, add or delete prompt: Delete prompt
I hope someone can help me. Thanks in advance.
If you really don't want all employees to see the toolbars, then you're taking on quite a bit of extra work. I would still consider this as an option, as you can apply permissions so that a user can only edit the content you allow, which means that users can take full advantage of Django CMS's built in functionality, which is great.
If you still don't want to take this route then you're going to have to build your own mini admin for your article model. Below I've quickly thrown together an idea for how you can approach this to hopefully help point you in the right direction.
First, your article view should be something like:
from django.views.generic import DetailView
from .models import Article
class ArticleView(DetailView):
context_object_name = 'article'
model = Article
template_name = 'path/to/article.html'
def get_context_data(self, **kwargs):
context = super(ArticleView, self).get_context_data(**kwargs)
context['show_controls'] = (self.request.user.is_authenticated() and
context[self.context_object_name].article == self.request.user)
return context
With the article template like:
<section>
{% if show_controls %}
<div class="controls">
Delete
Edit
</div>
{% endif %}
<article>
...
</article>
</section>
The path to delete view could be a confirm page like the Django admin. So you'd have a view like:
from django.contrib.auth.decorators import login_required
from django.core.exceptions import PermissionDenied
from django.shortcuts import get_object_or_404, redirect, render
from .models import Article
#login_required
def delete_article(request, article_pk):
if request.method == "POST":
article = get_object_or_404(Article, pk=article_pk)
if request.user != article.author:
raise PermissionDenied
article.delete()
return redirect('/redirect/url')
else:
context = {}
...
return render(request, 'path/to/confirm/delete.html', context)
With a template along the lines of:
<section>
<form method="POST">
{% csrf_token %}
<p>Are you sure you want to delete?</p>
<input type="submit" value="Delete">
</form>
</section>
You'd then create a similar setup for the edit page, navigate the user to a page that has a form where the fields can be amended and submitted etc.
Python 3.5.1
Django 1.10
Been tearing my hair out for hours on this, but have my Reverse Argument error pinned down to the actual problem.
When I try to open a form to edit a particular record in my model, it only opens a blank (unconnected) form. Using the same logic, I am able to delete a record, so I'm sure this is something stupid-simple. But I'm too many hours into this, so I would appreciate a lifeline.
From models.py
class CmsSpaces(models.Model):
sid = models.AutoField(db_column='SID', primary_key=True)
section = models.CharField(db_column='Section', max_length=5)
...Many more columns...
def __unicode__(self):
return self.name
def get_absolute_url(self):
return reverse('cms_spaces:space_edit', args = (self.sid), kwargs=None)
return reverse('cms_spaces:space_delete', args = (self.sid), kwargs=None)
return reverse('cms_spaces:space_new', args = None, kwargs = None)
class Meta:
managed = False
db_table = 'cms_spaces'
From views.py
def CmsSpaces_update(request, sid,
template_name='space_edit.html'):
space = get_object_or_404(CmsSpaces, sid=sid)
form = space_form(request.POST or None, instance=space)
if form.is_valid():
form.save()
return redirect('space_list')
return render(request, template_name, {'form':space_form})
def CmsSpaces_delete(request, sid,
template_name='space_delete.html'):
space = get_object_or_404(CmsSpaces, sid=sid)
if request.method=='POST':
space.delete()
return redirect('space_list')
return render(request, template_name, {'object':CmsSpaces})
From urls.py
from django.conf import settings
from django.conf.urls import include, url
from django.contrib import admin
from cms_spaces import views
urlpatterns = [
url(r'^space_list/$', views.CmsSpaces_list, name='space_list'),
url(r'^space_new/$', views.CmsSpaces_create, name='space_new'),
url(r'^space_edit/(?P<sid>[\w-]+)/$', views.CmsSpaces_update, name='space_edit'),
url(r'^space_delete/(?P<sid>[\w-]+)/$', views.CmsSpaces_delete, name='space_delete'),
]
From space_edit.html. When I enter the url directly for .../space_delete/12345, it does proceed to delete the record with sid=12345.
<form method="post">{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Add or Update Space" />
</form>
From space_list.html (which throws "Reverse for 'space_edit' with arguments '(10256,)' and keyword arguments '{}' not found. 0 pattern(s) tried: []") as an error. 10256 is in fact an sid for one of the records in the table. When I remove the links to space_edit and space_delete, it does render the list of records. Yes, there are many more lines of code that handle presentation, but this is the boiled down version, and it is "broken" even at this level.
<ul>
{% for CmsSpaces in space_list %}
<li>{{CmsSpaces.sid}} {{ CmsSpaces.section }} {{ CmsSpaces.space}} {{ CmsSpaces.lot }} {{ CmsSpaces.deednum}}
edit
delete
</li>
{% endfor %}
</ul>
All I want to do is be able to call and edit a record. If I can do that, I believe the list view will work, and then I'm well on my way to home free. The issue seems to be that even though it is correctly capturing the sid, it is failing to correctly pass it as an argument to the model. Help! Thanks.
EDIT
from django import forms
from django.forms import ModelForm
from masterdata.models import CmsSpaces
class space_form(ModelForm):
class Meta:
model = CmsSpaces
fields = [
'sid',
'section',
'space',
'lot',
'deednum',
'purchdate',
'notes',
'perpetualcare',
]
Okay... first off a big thanks to #knbk for getting me on track with the list issue.
Second, my initial diagnosis in the OP was wrong. The issue was in my CmsSpaces_update view. Boiled down, the view had the pieces to tie the form to the record, but no actual code to do so. slaps head
So because someone else may someday read this and wonder the same thing...
def CmsSpaces_update(request, sid,
template_name='templates/space_edit.html'):
space = get_object_or_404(CmsSpaces, sid=sid)
form = space_form(request.POST or None, instance=space)
if form.is_valid():
form.save()
return redirect('space_list')
ctx = {}
ctx['form'] = form
ctx['space'] = space
return render(request, template_name, ctx)
I am having problems with user authentication for my django site. I have a log-in screen that seems to work. When the user clicks log-in, I call the django.contrib.auth.login and it seems to work fine. However on subsequent pages have no knowledge that there is a user logged in. Example {% user.is_authenticated %} is false. There are also some menu functions that I want to be available for logged in users such as my-account and logout. Those functions are not available, except on the log-in page. Which is really strange.
This seems to be a user context problem. But I'm not sure how I am supposed to be passing a context around to ensure that my login is stable. Does anyone know at could be going on here? Any advice?
---------part of base.html------------
<!--- The following doesn't register even though I know I'm authenticated -->
{% if user.is_authenticated %}
<div id="menu">
<ul>
<li>My Customers</li>
<li>Customer Actions</li>
<li>My Account</li>
</ul>
</div>
{% endif %}
---------my views.py -----------------
# Should I be doing something to pass the user context here
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
c = Context({
'customer_list': customer_list,
})
t = loader.get_template(template)
return HttpResponse(t.render(cxt))
If you're using Django 1.3, you can use the render() shortcut, which automatically includes RequestContext for you.
from django.shortcuts import render
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
return render(request, "path_to/template.html",
{'customer_list': customer_list,})
In this case, you could go one step further, and use the generic ListView:
from django.views.generic import ListView
class CustomerList(Listview):
template_name = 'path_to/template.html'
queryset = Customer.objects.all().order_by('lastName')[:5]
Use a RequestContext.
As Daniel Suggested, use the RequestContext... or better, just use the render_to_response shortcut:
from django.template import RequestContext
from django.shortcuts import render_to_response
def customer_list(request):
customer_list = Customer.objects.all().order_by('lastName')[:5]
return render_to_response(
"path_to/template.html",
{'customer_list':customer_list,},
context_instance=RequestContext(request))