I'm working on a little app that allows you to save specific location information about places you've been. The issue I'm having is that clicking the submit button on the 'save new location' page doesn't seem to be doing much of anything. It redirects to .../locationlib/savenew/, which is supposed to be the url that saves the form input as a new model object, but both according to debugging print statements and what actually happens, that function is just never called. I've had success with other forms using django but this one seems to be tripping me up. Can someone give me an idea of what's going on here?
views.py
def new(request):
return render(request, 'locationlib/new.html')
def savenew(request):
print 'savenew called'
name = request.POST['name']
latitude = float(request.POST['latitude'])
longitude = float(request.POST['longitude'])
desc = request.POST['description']
user = User.objects.get(username=str(request.POST['user']))
print 'all variables set'
l = Location(
name=name,
longitude=longitude,
latitude=latitude,
custDescription=desc,
user=user,
)
print 'l defined'
l.save()
print 'l saved'
return HttpResponseRedirect(reverse('locationlib:detail', args=[l.id]))
new.html
<div id='new-location-form'>
<form action="{% url 'locationlib:savenew' %}" method="post">
{% csrf_token %}
Name: <input type='text' name='name' value='Place Name' required><br>
User: <input type='text' name='user' value='User' required><br>
Longitude: <input type='text' name='longitude' value="Longitude Coordinate" required><br>
Latitude: <input type='text' name='latitude' value='Latitude Coordinate' required><br>
Description: <textarea name='description'>Description of Place</textarea><br>
<input type="submit" value="Save">
</form>
</div>
urls.py
urlpatterns = patterns( '',
...
url(r'new/', views.new, name='new'),
url(r'^savenew/', views.savenew, name='savenew'),
)
Your first URL pattern, new, is not anchored to the start of the string. That means that it matches anything that ends with "new" - and that includes "savenew". So your request for "savenew" is being caught by that pattern, and being sent to the new view.
Just put a ^ character at the front, as you have done with the other pattern.
try to use Modelforms
forms.py:
from django.forms import ModelForm
from myapp.models import Location
# Create the form class.
class LocationForm(ModelForm):
class Meta:
model = Location
view.py
def savenew(request):
if request.method == 'POST':
form = LocationForm(request.POST)
if form.is_valid():
new=form.save()
return HttpResponseRedirect(reverse(reverse('locationlib:detail', args=[new.id])))
return render(request,'reservetion/sign.html',{'form': form})
else:
form = SignForm()
return render(request, 'reservetion/sign.html',{'form': form})
new.html
<form action="{% url 'locationlib:savenew' %}" method="post">
{% csrf_token %}
{{ form}}
</form>
Related
I'm getting this error when submit:
CSRF verification failed. Request aborted.
I've got this far following the documentation, but I don't fully understand it and it's definitely wrong. I just want to take a query word from my search box(form) and pass it to a python script as an argument. I'm new to Django and getting stuck on the easiest things.
In models.py:
class QueryForm(forms.Form):
query = forms.CharField(label='query',max_length=100)
I added this line to my urls.py
url(r'^results/$', 'tweemo.views.results'),
On my homepage where my search box is I have this code for my form:
<form action="/home/results/" method="post">
<label for="query">Search:</label>
<input id="query" type="text" name="query" value="{{ current_query }}">
<input type="submit" value="ok">
</form>
In views.py I added these two functions:
def get_query(request):
if request.method == 'POST':
form = QueryForm(request.POST)
if form.isvalid():
return HttpResponseRedirect('/thanks/')
else:
form = QueryForm()
return render(request, 'results.html', {'form': form})
def results(request):
return render_to_response('results.html',{'here':TwitterStream.objects.all() })
MY results.html contains just this:
<form action="/home/results/" method="post">
{% csrf_token %}
{{ form }}
<input type="submit" value="Submit"/>
</form>
You must just add the {% csrf_token %} tag inside EVERY <form></form> tag which has method to be post in your template.
So the below markup should be corrected:
<form action="/home/results/" method="post">
{% csrf_token %}
<label for="query">Search:</label>
<input id="query" type="text" name="query" value="{{ current_query }}">
<input type="submit" value="ok">
</form>
Well the problem is that you are not passing the csrf token to the form , you need to pass the csrf token to the render function in order for it to be applied in the form . To accomplish this you need to associate the csrf token to the request.
def get_query(request):
if request.method == 'POST':
form = QueryForm(request.POST)
if form.isvalid():
return HttpResponseRedirect('/thanks/')
else:
form = QueryForm()
args = {}
args.update(csrf(request))
args['form'] = form
return render_to_response('results.html', args)
def results(request):
return render_to_response('results.html',{'here':TwitterStream.objects.all() })
I want the user to be redirected to the last page after successful login whenever the user is prompted to login. Currently, it just redirects to the index page. I know this question has been asked several times and have also tried multiple ways of doing it but it doesnt work. I've also tried following this Django: Redirect to previous page after login
and it doesn't work because most of the stuff is already deprecated in 1.6. So I've the following code now and it gives me this 'url' requires a non-empty first argument. The syntax changed in Django 1.5, see the docs.
on line
129 : <h4><a class="writebtn"href=" {% url django.contrib.auth.views.login %} ?next={{request.path}}">Write Review</a></h4>
This is what i've so far
urls.py
url(r'^login/$', 'django.contrib.auth.views.login', {'template_name': 'meddy1/login.html'}, name="login")
login.html
<form method="post" action="{% url 'django.contrib.auth.views.login' %}"> {% csrf_token %}
<input type="text" class="form-control" id="username" placeholder="Username" name="username">
<input type="password" class="form-control" id="inputPassword" placeholder="Password" name="password">
<button class="btn btn-primary" type="submit" name="submit" id="ss-submit">LOG IN</button>
<input type="hidden" name="next" value="{{ next }}" />
</form>
I've this in my template to see if the user is authenticated to write a review - if not logged in it renders the login page but then it redirects to the index page
{% if user.is_authenticated and reviewed == False %}
<h4><a class="writebtn"href="/usercontent/{{doctor.id}}/">Write Review</a></h4>
{% elif user.is_authenticated and reviewed == True %}
<h4><a class="writebtn"href="">Already Reviewed!</a></h4>
{% else %}
<h4><a class="writebtn"href="{% url django.contrib.auth.views.login %}?next={{request.path}}">Write Review</a></h4>
{% endif %}
views.py
#login_required
def addContent(request, id):
d = getVariables(request)
doctor = Doctor.objects.get(id=id)
params = {}
params.update(csrf(request))
if request.user.is_authenticated():
user = request.user
ds = DoctorSeeker.objects.get(user=user)
d['doctorseeker'] = ds
if request.method == "POST":
form = UserContentForm(request.POST)
if form.is_valid():
time = form.cleaned_data['time']
comment = form.cleaned_data['comment']
if request.POST.get('Like') == 'Like':
con = UserContent(time=time, comment = comment, liked = True, disliked = False, doctor_id = doctor.id, user_id = request.user.id)
doctor.likes += 1
doctor.netlikes = doctor.likes - doctor.dislikes
doctor.save()
con.save()
elif request.POST.get('Like') == 'Dislike':
con = UserContent(time=time, comment = comment, liked = False, disliked = True, doctor_id = doctor.id, user_id = request.user.id)
doctor.dislikes +=1
doctor.netlikes = doctor.likes - doctor.dislikes
doctor.save()
con.save()
url = '/docprofile/%s' % str(doctor.id)
return HttpResponseRedirect(url)
else:
form = UserContentForm()
d.update({'doctor': doctor, 'UGC': UserContent.objects.all(),
'form': form })
return render(request, 'meddy1/usercontent.html',d)
For the following template code to work:
Link
The following must be in place:
You must have a reference to django.contrib.auth.views.login in your `urls.py:
url(r'^login/$', 'django.contrib.auth.views.login')
You must have a template that resolves to registration/login.html or you can specify a specific template path (which must exist) in the your url conf:
url(r'^login/$',
'django.contrib.auth.views.login',
{'template_name': '<your template>'}
)
In your settings.py, you must have:
TEMPLATE_CONTEXT_PROCESSORS = (
'django.core.context_processors.request',
...
)
This automatically adds the request variable gets to your template context.
You must render your template with a RequestContext. There are many ways to do that, but the most simple is:
from django.shortcuts import render
def view(request):
...
return render(request, '<template>', {<other context variables>})
Using a RequestContext causes the django.core.context_processors.request context processor to run.
I am working with Django forms and for some reason, this form will not validate! It submits alright, or at least the runserver shows an http post response with code 200 (ok). For some reason though, my form will not pass the is_valid test!
views.py:
def new_show(request):
if request.method == 'POST':
img_form = ImageForm(request.POST, request.FILES)
show_form = NewShowForm(request.POST)
if show_form.is_valid():
new_Show = Show()
new_Show.title=show_form.cleaned_data['title']
new_Show.body=show_form.cleaned_data['body']
new_Show.pub_date=timezone.now()
new_Show.location=show_form.cleaned_data['location']
new_Show.time=show_form.cleaned_data['time']
new_Show.save()
if img_form.is_valid():
image=Image(image=request.FILES['imageFile'])
new_Show.image_set.add(image)
return HttpResponseRedirect(reverse('shows'))
else:
return HttpResponseRedirect(reverse('shows'))
else:
show_form = NewShowForm()
img_form = ImageForm()
return render_to_response(
'shows/new_show.html',
{'show_form': show_form, 'img_form': img_form},
context_instance=RequestContext(request)
)
Here is my template snippet:
<form action="{% url "new_show" %}" method="post" enctype="multipart/form-data">
{% csrf_token %}
<p>{{ show_form.non_field_errors }}</p>
<p>
<label for="title">Title:</label>
<input type="text" name="title"/>
</p>
<p>
<label for="body">Body:</label>
<textarea type="text" name="body"> </textarea>
</p>
<p>
<label for="location">Location:</label>
<input type="text" name="location"/>
</p>
<p>
<label for="time">Date:</label>
<input type="text" id="time" maxlength="25" size="25" name="time"><img src="{{ STATIC_URL }}../../static/cal.gif" width="16" height="16" border="0" alt="Pick a date">
</p>
<!-- Upload Form. Note enctype attribute! -->
{% csrf_token %}
<p>{{ img_form.non_field_errors }}</p>
<p>{{ img_form.imageFile.label_tag }}</p>
<p>
{{ img_form.imageFile.errors }}
{{ img_form.imageFile }}
</p>
<p><input type="submit" value="Add Upcoming Show"></input></p>
</form>
Here is my form Class:
class NewShowForm(forms.Form):
title=forms.CharField()
body=forms.CharField(widget=forms.TextArea)
location=forms.CharField()
time=forms.DateTimeField(required=True)
class ImageForm(forms.Form):
imageFile = forms.FileField(required=False, label='Select an Image')
Please help me!
If new_Show is a model, why not create a ModelForm instead of forms.Form?
So, instead of
class NewShowForm(forms.Form):
title=forms.CharField()
body=forms.CharField(widget=forms.TextArea)
location=forms.CharField()
time=forms.DateTimeField(required=True)
class ImageForm(forms.Form):
imageFile = forms.FileField(required=False, label='Select an Image')
why not using,
from django.forms import ModelForm
class NewShowForm(ModelForm):
class Meta:
model = NewShow
class ImageForm(ModelForm):
class Meta:
model = Image
?
Using ModelForm will ensure that form validation meets that of model. Moreover, it can cut off your code (especially line 6 to 11).
It will help to add these two lines to your view before if is_valid() to see the errors it's giving:
if request.method == 'POST':
img_form = ImageForm(request.POST, request.FILES)
show_form = NewShowForm(request.POST)
print(form.is_valid())
print(form.errors)
if show_form.is_valid():
You can paste the errors here and we can see what's the issue
Since you've put 2 Django forms together under one HTML form tag, when you submit the form on the front-end you're sending an extra field through request.POST that your NewShowForm doesn't have. If you combine both forms into a single Django form, you should be able to get this to work.
This is my form:
from django import forms
class UploadFileForm(forms.Form):
titl = forms.CharField(max_length=50)
ffile = forms.FileField()
This is my views.py file:
def handle_uploaded_file(file_path):
print "handle_uploaded_file"
dest = open(file_path.name,"wb")
for chunk in file_path.chunks():
dest.write(chunk)
dest.close()
def handle_upload(request):
c = {}
c.update(csrf(request))
if request.method == "POST":
form = UploadFileForm(request.POST)
if form.is_valid():
handle_uploaded_file(request.FILES["ffile"])
return HttpResponseRedirect("/thanks")
else:
form = UploadFileForm()
c.update({"form":form})
return render_to_response("upload.html",c)
And this is the content of upload.html:
<form enctype="multipart/form-data" method="post" action="/handle_upload/">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Upload it"/>
</form>
Whenever I try to submit the form, I get a "This field is required" for the ffile field. What am I doing wrong? Just to mention, I am uploading a file each time.
Just for future reference. I had the same error, though I included request.FILES in form initialization. The problem was in the template: I forgot to add enctype="multipart/form-data" attribute to the <form> tag.
form = UploadFileForm(request.POST, request.FILES)
If you have included request.FILES and added the enctype="multipart/form-data", but are still seeing this error, it could be you are not declaring the <input> correctly.
For example, if explicitly declare the input html in your template like:
<input type="file" value="Upload CSV File" />
You may not be passing the expected input id or name attributes of the input form element.
Be sure that your template is using the form element tag, i.e. {{ form.file }},
which django will then render as: <input id="id_file" name="file" type="file" required=""> on the page.
I have form with one input for email and two submit buttons to subscribe and unsubscribe from newsletter:
<form action="" method="post">
{{ form_newsletter }}
<input type="submit" name="newsletter_sub" value="Subscribe" />
<input type="submit" name="newsletter_unsub" value="Unsubscribe" />
</form>
I have also class form:
class NewsletterForm(forms.ModelForm):
class Meta:
model = Newsletter
fields = ('email',)
I must write my own clean_email method and I need to know by which button was form submited. But the value of submit buttons aren't in self.cleaned_data dictionary.
Could I get values of buttons otherwise?
Eg:
if 'newsletter_sub' in request.POST:
# do subscribe
elif 'newsletter_unsub' in request.POST:
# do unsubscribe
You can use self.data in the clean_email method to access the POST data before validation. It should contain a key called newsletter_sub or newsletter_unsub depending on which button was pressed.
# in the context of a django.forms form
def clean(self):
if 'newsletter_sub' in self.data:
# do subscribe
elif 'newsletter_unsub' in self.data:
# do unsubscribe
You can also do like this,
<form method='POST'>
{{form1.as_p}}
<button type="submit" name="btnform1">Save Changes</button>
</form>
<form method='POST'>
{{form2.as_p}}
<button type="submit" name="btnform2">Save Changes</button>
</form>
CODE
if request.method=='POST' and 'btnform1' in request.POST:
do something...
if request.method=='POST' and 'btnform2' in request.POST:
do something...
one url to the same view!
like so!
urls.py
url(r'^$', views.landing.as_view(), name = 'landing'),
views.py
class landing(View):
template_name = '/home.html'
form_class1 = forms.pynamehere1
form_class2 = forms.pynamehere2
def get(self, request):
form1 = self.form_class1(None)
form2 = self.form_class2(None)
return render(request, self.template_name, { 'register':form1, 'login':form2,})
def post(self, request):
if request.method=='POST' and 'htmlsubmitbutton1' in request.POST:
## do what ever you want to do for first function ####
if request.method=='POST' and 'htmlsubmitbutton2' in request.POST:
## do what ever you want to do for second function ####
## return def post###
return render(request, self.template_name, {'form':form,})
/home.html
<!-- #### form 1 #### -->
<form action="" method="POST" >
{% csrf_token %}
{{ register.as_p }}
<button type="submit" name="htmlsubmitbutton1">Login</button>
</form>
<!--#### form 2 #### -->
<form action="" method="POST" >
{% csrf_token %}
{{ login.as_p }}
<button type="submit" name="htmlsubmitbutton2">Login</button>
</form>
It's an old question now, nevertheless I had the same issue and found a solution that works for me: I wrote MultiRedirectMixin.
from django.http import HttpResponseRedirect
class MultiRedirectMixin(object):
"""
A mixin that supports submit-specific success redirection.
Either specify one success_url, or provide dict with names of
submit actions given in template as keys
Example:
In template:
<input type="submit" name="create_new" value="Create"/>
<input type="submit" name="delete" value="Delete"/>
View:
MyMultiSubmitView(MultiRedirectMixin, forms.FormView):
success_urls = {"create_new": reverse_lazy('create'),
"delete": reverse_lazy('delete')}
"""
success_urls = {}
def form_valid(self, form):
""" Form is valid: Pick the url and redirect.
"""
for name in self.success_urls:
if name in form.data:
self.success_url = self.success_urls[name]
break
return HttpResponseRedirect(self.get_success_url())
def get_success_url(self):
"""
Returns the supplied success URL.
"""
if self.success_url:
# Forcing possible reverse_lazy evaluation
url = force_text(self.success_url)
else:
raise ImproperlyConfigured(
_("No URL to redirect to. Provide a success_url."))
return url
I know this is old, but some of the answers are, to say the least, brief, and they do not address a common case where the form is not a django form.
This solution was inspired by this blog post. It relies on using a view class that is derived from django.views.generic.edit.FormMixin, e.g. CreateView, UpdateView or DeleteView. These provide the get_success_url method which exposes the button name in request
html
<html>
<body>
<form method="post">
<div>
<label> <input type="radio" name="select-type" value="A">Type A</label>
</div>
<div>
<label> <input type="radio" name="select-type" value="B">Type B</label>
</div>
<div>
<input type="submit" value="Use selected">
</div>
<div>
<input type="submit" name="no-selection" value="None of the above">
</div>
</form>
</body>
</html>
views.py
from django.views.generic import UpdateView
class GetType(UpdateView):
def get(self, request):
return render(request, 'get_type.html', {})
def post(self, request):
button = self.get_success_url()
print(button)
def get_success_url(self):
if 'no-selection' in self.request.POST:
return 'none selected'
return ''