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Django: How can I take the choice in drop-down list and redirect to another page?

I'm currently just learning Django and I'm doing electronic grade book. I have tried everything, have read all the documentation, but nothing helps. It seems I miss a simple logic somewhere. I need to make two pages:
The first one "teacher_interface" is a simple inteface for the teacher with just one drop-down list, teacher chooses the necessary class (i.e 1C, 2B, 4C) and the button "Students", which should somehow take the chosen class from drop-down list input and redirect to the second page "class_students".
The second "class_students" is alike the "teacher_interface", but with the table of students of the chosen class.
I have the One-to-many relation between classes Student and Class:
Firstly, I tried redirecting from "teacher_interface" to "class_students", using in template:
{% url "name" %}
Parts of code: 1) models.py https://dpaste.org/eqxm 2) urls.py https://dpaste.org/eUEO 3) views.py https://dpaste.org/ap8D#L 4) template teacher_interface.html https://dpaste.org/v4m9 5) template class_students.html https://dpaste.org/0gXK
But it shows me: Reverse for 'class_students' with no arguments not found. 1 pattern(s) tried: ['school/teacher/(?P<class_id>[0-9]+)/class/$']
I tried everything, but nothing helped, this and the similar: Django - getting Error "Reverse for 'detail' with no arguments not found. 1 pattern(s) tried:" when using {% url "music:fav" %} I understood maybe this two options of redirect will not work in my case:
{% url 'class_students' class.id %}
{% url 'class_students' class_id %}
I also don't know if it's possible to do on the same page.
So I decided to redirect using redirect from django.shortcuts. I changed my teacher_interface view, so that it took the id of the chosen by the teacher class if request method is POST and redirected. I also made this change in my template "teacher_interface.html":
from
action="{% url 'class_students' %}"
to
action=""
Changed view:
def teacher_interface(request):
class_queryset = Class.objects.order_by("class_number", "group")
class_id = None
if request.method == "POST":
class_id = Class.objects.get("id")
return redirect("class_students", class_id)
context = {
"class_queryset": class_queryset,
"class_id": class_id,
}
return render(request, "teacher_interface.html", context)
But when I choose the class and click the "Students" button, it shows me: Cannot resolve keyword 'i' into field. Choices are: class_number, curriculum, discipline, group, id, student, task, type_of_class, type_of_class_id. Id is certainly is a key, but it tries to resolve only "i".
I tried/read everything here, but nothing works.
I even wrote the default like this:
class_id = Class.objects.get("id", "default")
I am sure I just don't understand properly how to get teacher's choice, pass it to another or the same function and redirect, saving this information. I will be really grateful for you help, even if you just advise what I can read to figure it out.

Ok, you are missing some basic conpects.
on your views.py
def teacher_interface(request):
class_queryset = Class.objects.order_by("class_number", "group")
context = {
"class_queryset": class_queryset,
}
return render(request, "teacher_interface.html", context)
this is correct, you will pass you query to your template
on your template change some things to look like this:
<form method="POST" >{% csrf_token %}
<select name="input1">
{% for class in class_queryset %}
<option value="{{ class.id }}">{{ class }}</option>
{% endfor %}
</select>
<input type="submit" value="Students"/>
</form>
then you need to change your teacher_interface view:
You need to import redirect on your views.py
def teacher_interface(request):
class_queryset = Class.objects.order_by("class_number", "group")
context = {
"class_queryset": class_queryset,
}
if request.method == 'POST':
class_id = request.POST.get('input1') # I'm not sure if this will get the {{class.id}} value, if don't, print(request.POST.get) and check how to get the value
return redirect('class_students', class_id=class_id) # will make a get request on the class_students view
return render(request, "teacher_interface.html", context)
def class_students(request, class_id):
# the parameter need to be 'class_id' because this is what you put on your urls '<int:class_id>', if possible, remove that /class.
# ADD CLASS ID AS PARAMETER, THAT WILL ENABLE YOU TO ACESS AN SPECIFIC CLASS
# Import get_object_or_404 (google it and you will find easily)
class = get_object_or_404(Class, pk=class_id) # this avoid internal server error.
# pass your class on the context
return render(request, "class_students.html")

Thwarting form double-submission through server side tokens (Django)

I am trying to implement a server-side check to prevent users from double-submitting my forms (Django web app).
One technique I'm trying is:
1) When the form is created, save a unique ID in the session, plus pass the unique ID value into the template as well.
2) When the form is submitted, pop the unique ID from the session, and compare it to the same unique ID retrieved from the form.
3) If the values are the same, allow processing, otherwise not.
These SO answers contributed in me formulating this.
Here's a quick look at my generalized code:
def my_view(request):
if request.method == 'POST':
secret_key_from_form = request.POST.get('sk','0')
secret_key_from_session = request.session.pop('secret_key','1')
if secret_key_from_form != secret_key_from_session:
return render(request,"404.html",{})
else:
# process the form normally
form = MyForm(request.POST,request.FILES)
if form.is_valid():
# do something
else:
# do something else
else:
f = MyForm()
secret_key = uuid.uuid4()
request.session["secret_key"] = secret_key
request.session.modified = True
return render(request,"my_form.html",{'form':f,'sk':secret_key})
And here's a sample template:
<form action="{% url 'my_view' %}" method="POST" enctype="multipart/form-data">
{% csrf_token %}
<input type="hidden" name="sk" value="{{ sk }}">
{{ form.my_data }}
<button type="submit">OK</button>
</form>
This set up has failed to stop double-submission.
I.e., one can go on a click frenzy and still end up submitting tons of copies. Moreover, if I print secret_key_from_form and secret_key_from_session, I see them being printed multiple times, even though secret_key_from_session should have popped after the first attempt.
What doesn't this work? And how do I fix it?
UPDATE: when I use redis cache to save the value of the special key, this arrangement works perfectly. Therefore, it seems the culprit is me being unable to update request.session values (even with trying request.session.modified=True). I'm open to suggestions vis-a-vis what could be going wrong.
Note that this question specifically deals with a server-side solution to double-submission - my JS measures are separate and exclusive to this question.

You might just need request.session.modified = True. If you want to make sure that the session is deleting you can use del too.
secret_key_from_session = request.session.get('secret_key','1')
del request.session['secret_key']
request.session.modified = True

I couldn't figure out what caused the problem, but via substituting Redis cache for every request.session call, I was able to get my desired results. I'm open to suggestions.

Changing Id of django form in for loop

I'm stuck in my code. Need help.
This is my front end. I am rendering forms stored in "form_list".
The problem is that the forms stored are of same type and thus produce input fields with same "id" and same "name".
This is my view:-
#login_required
def VideoLinkView(request):
"""view to save the video links """
current_form_list = []
current_form = None
if request.method == 'GET':
vl = VideoLink.objects.filter(company=CompanyModel.objects.get(owner=request.user))
for link in vl:
current_form = VideoLinkForm(link.__dict__)
current_form_list.append(current_form)
return render(request, "premium/video_link.html", context={'form_list':current_form_list})
This is my html template :-
{% for form in form_list %}
<div class="form-group">
<label for="id_video_link">Video Link:</label>
{{ form.video_link }}
</div>
{% endfor %}
How can I create different "id" and different "name" in each iteration of for loop's input tag, automatically without having knowledge of no form stored in form_list.
I tried {{ forloop.counter}} it didn't worked, perhaps I made some mistake. Also, raw python don't work in template.
Thanks in Advance.

The way you are creating your forms is wrong in two ways. Firstly, the first positional argument is for the values submitted by the user; passing this arg triggers validation, among other things. If you are passing values for display to prepopulate the form, you must use the initial kwarg:
current_form = VideoLinkForm(initial={...dict_of_values...})
However, even that is not correct for your use case here. link is a model instance; you should use the instance kwarg:
current_form = VideoLinkForm(instance=link)
Now, to solve the problem you asked, you could just pass a prefix as well as I originally recommended:
for i, link in enumerate(vl):
current_form = VideoLinkForm(instance=link, prefix="link{}".format(i))
However, now that you have shown all the details, we can see that this is not the best approach. You have a queryset; so you should simply use a model formset.
from django.forms import modelformset_factory
VideoLinkFormSet = modelformset_factory(VideoLink, form=VideoLinkForm, queryset=vl)
current_form_list = VideoLinkFormSet()

Django database inquiries. Finding duplicates in the database and stopping them from being submitted

I'm trying to go inside of my database and see if a certain submission has been submitted before and if it has stop it from being submitted again. Currently I have this bit of code testing each field of the form (obviously changes per field but i thought I'd only show one field for simplicity)
if request.method == 'POST':
Cashtexts.objects.filter(cashTexts=request.POST['cashTexts']).exists() == False:
Then if it does not exist it goes on and save the submissions into the database. If it returns true it goes to an else statement that tells the user that what they entered has already been entered. previously this type of thing worked but I changed some variable names and now it stopped working. I've been browsing the code over a few time so I thought maybe something was fundamentally wrong with this type of filter but it makes sense in my head.
def submit_win(request):
if request.method == 'POST':
if Cashtexts.objects.filter(cashTexts=request.POST['cashTexts']).exists() or Cashtexts.objects.filter(superPoints=request.POST['superPoints']).exists() or Cashtexts.objects.filter(varolo= request.POST['varolo']).exists() or Cashtexts.objects.filter(swagbucks = request.POST['swagbucks']).exists() or Cashtexts.objects.filter(neobux = request.POST['neobux']).exists() or Cashtexts.objects.filter(topline=request.POST['topline']).exists() or Cashtexts.objects.filter(Paidviewpoint=request.POST['Paidviewpoint']).exists() or Cashtexts.objects.filter(cashcrate=request.POST['cashcrate']).exists() == False:
form = CashtextsForm(request.POST)
if form.is_valid():
form.save()
ref_create = Cashtexts.objects.filter(cashTexts=request.POST['cashTexts'])
return render_to_response('submitted_page.html', {'ref_create': ref_create})
else: #Error message reading wither you didnt insert codes or you enter the same code twice
error = 'That code has already been submitted'
ref_create = CashtextsForm()
return render_to_response('submit.html', {'ref_create': ref_create,'error':error}, context_instance=RequestContext(request))
else: #displays the page when the user asks to go to the submit page
ref_create = CashtextsForm()
return render_to_response('submit.html', {'ref_create': ref_create}, context_instance=RequestContext(request))
Also, I turned Cashtexts.objects.filter(cashTexts=request.POST['cashTexts']) intoa variable and passed it to my template to see what it was returning and it was returning objects that should have told the statement True. but it just seemed to ignore them and submit them anyway.
I suppose i could delete all the previous objects that match the thing they entered upon submission but that makes it less secure and could lead to users submitting over and over again thinking that they are getting it more in there. I'd rather just stop it before it happens.

You can just set the definition of your cashTexts field to unique in the models definition:
def Cashtexts(models.Model):
name = CharField(max_length=50, unique = True)
Just set the "unique" argument to true for each field you want to be unique and you're done. The way Django's forms API is working, the form fields will take over all the necessary error handling.
With the model fields set to unique where necessary you can have this much easier:
def view(request):
if request.method == 'POST':
form = CashtextsForm(request.POST)
""" the next line is going to check uniqueness for each
of the models fields where you have set unique = True
If the field is not unique, it s field will have an
attribute error, which you can then render to the template """
if form.is_valid():
form.save()
else:
form = CashtextsForm()
context = RequestContext(request)
# now i pass the form to the template as 'ref_create'
context.update( {'ref_create':form} )
return render_to_response('submit.html',context)
Then you simply render that form in your template (either with form.as_p(), form.as_table() or with some custom code like this):
{% for field in ref_create %}
{{ field.label_tag }}
{{ field }}
{{ field.errors }}
{% endfor %}
Each field, set to unique in the model definition, which which would not be unique in the database any more if your form is saved, will now have an error message saying something like "A Cashtext with this name allready exists".
Customizing you error messages can be done in two ways: 1: customize the forms (i do not recommend that, until you have better understandings of how Django works) 2: instead of {{ field.errors }} write something like this:
{% if field.errors %}
The Cashtext field {{field.name}} value should be unique.
{% endif %}

POSTing forms in Django's admin interface

I'm writing a Django admin action to mass e-mail contacts. The action is defined as follows:
def email_selected(self,request,queryset):
rep_list = []
for each in queryset:
reps = CorporatePerson.objects.filter(company_id = Company.objects.get(name=each.name))
contact_reps = reps.filter(is_contact=True)
for rep in contact_reps:
rep_list.append(rep)
return email_form(request,queryset,rep_list)
email_form exists as a view and fills a template with this code:
def email_form(request,queryset,rep_list):
if request.method == 'POST':
form = EmailForm(request.POST)
if form.is_valid():
cd = form.cleaned_data
send_mail(
cd['subject'],
cd['message'],
cd.get('email','noreply#localboast'),['redacted#email.com'],
)
return HttpResponseRedirect('thanks')
else:
form = EmailForm()
return render_to_response('corpware/admin/email-form.html',{'form':form,})
and the template exists as follows:
<body>
<form action="/process_mail/" method="post">
<table>
{{ form.as_table }}
</table>
<input type = "submit" value = "Submit">
</form>
</body>
/process_mail/ is hardlinked to another view in urls.py - which is a problem. I'd really like it so that I don't have to use <form action="/process_mail/" method="post"> but unfortunately I can't seem to POST the user inputs to the view handler without the admin interface for the model being reloaded in it's place (When I hit the submit button with , the administration interface appears, which I don't want.)
Is there a way that I could make the form POST to itself (<form action="" method="post">) so that I can handle inputs received in email_form? Trying to handle inputs with extraneous URLs and unneeded functions bothers me, as I'm hardcoding URLs to work with the code.

You can use django's inbuilt url tag to avoid hardcoding links. see...
http://docs.djangoproject.com/en/dev/ref/templates/builtins/#url
Chances are you'd be better off setting up a mass mailer to be triggered off by a cron job rather than on the post.
Check out the answer I posted here
Django scheduled jobs
Also if you insist on triggering the email_send function on a view update perhaps look at
http://docs.djangoproject.com/en/dev/topics/signals/