I've created a form which by submit uploads an item to the database. The problem is that if I press f5 it'll submit the form again, because of the URL is now different.
I have these two url patterns
urlpatterns = [
url(r'(?i)^CMS/$', views.CMS, name='CMS'),
url(r'^createItem/$', views.createItem, name='createItem')
]
and my view looks like this
def CMS(request):
form = itemCreateForm()
context = {
'form' : form,
'message' : 'Content Manage Site'
}
return render(request, 'CMS.html', context)
def createItem(request):
f = itemCreateForm(request.POST)
if f.is_valid():
f.save()
pass
form = itemCreateForm()
context = {
'form' : form,
'message' : 'ItemCreated!'
}
return render(request, 'CMS.html', context)
the CMS.html
{% if message %}
{{ message }}
{% endif %}
<div class='newItemFields'>
<form action="{% url 'kar:createItem' %}" method="POST">
{% csrf_token %}
{{ form.as_p }}
<input type="submit">
</form>
</div>
my form
class itemCreateForm(ModelForm):
class Meta:
model = item
fields = ['name', 'type', 'price']
I start at homepage/CMS/ and fill in the form and press submit, and view function createItem runs and creates and saves the object in the database. And sends the user to homepage/CMS/createItem. And now everytime the user press f5 the createItem function will run again and insert another object into the database with the same values as the previous one, even though the input fields are empty (can't wrap my head around that).
I also twice write form = itemCreateForm() which I believe is dubious?
What I'd like to do is after createItem is run, it should send the user back to homepage/CMS/ and not homepage/CMS/createItem. Would that be the proper way to do it? Or is there a smart way of doing this.
At the end of your createItem function, you are rendering HTML of the page rather than redirecting. Instead, you need to do
return HttpResponseRedirect(reverse('kar:index'))
You will need to import HttpResponseRedirect and reverse which is used to resolve the URL through its name.
Check this out: https://docs.djangoproject.com/en/1.10/topics/forms/#the-view
What I'd like to do is after createItem is run, it should send the
user back to homepage/CMS/ and not homepage/CMS/createItem. Would that
be the proper way to do it? Or is there a smart way of doing this.
That would indeed be the proper and smart way to do it. Have one view handle both GET and POST and then redirect after successful form submission. This ensures that the user can't resubmit the form merely by refreshing. And you address your concern about repeating your code.
urlpatterns = [
url(r'(?i)^$', views.index, name='index'),
url(r'^createItem/$', views.createItem, name='createItem')
]
Then combine your views
def createItem(request):
if request.method == 'POST':
f = itemCreateForm(request.POST)
if f.is_valid():
f.save()
return HttpResponseRedirect('/homepage/CMS/')
else :
form = itemCreateForm()
context = {
'form' : form,
'message' : 'Content Manage Site'
}
return render(request, 'CMS.html', context)
Note that the code is now shorter, it gives proper feedback to the user when the form is not valid. And you can't refresh to submit the for twice. We need a small change to the template
<div class='newItemFields'>
<form action=method="POST">
{% csrf_token %}
{{ form.as_p }}
<input type="submit">
</form>
</div>
The message display part isn't needed anymore
Related
i created a view + form that creates two widgets + a button for a user. One to select a choice and another to type something in. Now i want to redirect the user after the clicking the button to another webpage displaying his input. (Generally i want to know how to access the userinput and further use it).
This is my form:
class Eingabefeld(forms.Form):
eingabefeld = forms.CharField(label="Flight Number",max_length=20)
a = Auswahlmoeglichkeiten.objects.all()
flughafenname = forms.ModelChoiceField(label="Target Airport",queryset=a,empty_label="-------")
source = forms.CharField(
max_length=50,
widget=forms.HiddenInput(),
required=False
)
This is my views.py:
def get_eingabe(request):
log = logging.getLogger(__name__)
if request.method =="POST":
eingabe = Eingabefeld(request.POST)
log.warn(eingabe)
if eingabe.is_valid():
return HttpResponseRedirect("answerrequest")
else:
eingabe = Eingabefeld()
return render(request, "app_one/labels.html", {"eingabe": eingabe})
def answerrequestseite(request):
return render(request, "app_one/answerrequest.html")
and this is my html ( the included html in this one is just for layout):
<head>
<title>Home</title>
</head>
<form method="post" novalidate>
{% csrf_token %}
{% include "app_one/bootstrap_layout2.html" with form=eingabe %}
<div class="row">
<div class="col-sm-5"></div>
<div class="col-sm-2">
<button type="submit" class="btn btn-primary btn-block">Let's Go!</button>
</div>
<div class="col-sm-5"></div>
</div>
</form>
So basically when opening my webpage "get_eingabe" gets called, and the template gets rendered, now when clicking the button the input is validated and after successfull validation a different URL is opened which will trigger the method "answerrequestseite". Now how do i pass the userinput (eingabefeld and flughafenname) into the other method which will render the template for the second URL?
I read alot about using "request.GET" but i am not quite sure where exactly to place it and how.
After if eingabe.is_valid(): create some variable containing the values you want.
then in you redirect you need to pass those values as get argument like:
your_url/?id=123
Then you can retrieve your variable in your views.py via
request.GET.get('id')
But in your case, you don't want to pass simple id, you want to pass user_input.
One way will be to sanitize this input to make it url compatible.
Otherwise the more flexible solution is to store the values in the session.
Session (via cookie)
# views.py
# Set the session variable
request.session['you_variable_name_here'] = 'the value'
# Retrieve the session variable
var = request.session.get['you_variable_name_here']
https://docs.djangoproject.com/en/2.2/topics/http/sessions/
For your exemple in the first view:
if eingabe.is_valid():
eingabefeld = eingabe.cleaned_data.get('eingabefeld')
flughafenname = eingabe.cleaned_data.get('flughafenname')
request.session['eingabefeld'] = eingabefeld
request.session['flughafenname'] = flughafenname.pk
return HttpResponseRedirect("answerrequest")
In the second view:
def answerrequestseite(request):
eingabefeld = request.session.get('eingabefeld')
flughafenname_pk = request.session.get('flughafenname')
flughafenname = YourFlughafennameModel.objects.get(pk=flughafenname_pk)
return render(request, "app_one/answerrequest.html",{'eingabefeld':eingabefeld,'flughafenname':flughafenname})
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 am new to django forms and Crispy Forms. I have some simple forms in a little forum Im developing. I think I don't need to use the %crispy% tag. I only need the form|crispy filter. However, I don't know why they don't render the error messages.
Also, if I want to customize the error messages (they must be in spanish), do I need to use the %crispy% tag or is it possible to do this with the |crispy filter?
Anyway, here is one of my forms:
from django import forms
from django.forms import Textarea
class FormNuevoVideo(forms.Form):
url = forms.URLField(initial='http://', max_length=250)
titulo = forms.CharField(max_length=150)
descripcion = forms.CharField(
help_text="...",
widget=Textarea(attrs={'rows': 3, 'data-maxlength': 500}))
Here is the view:
#login_required
def nuevo_video(request, slug):
template = 'videos/nuevo.html'
tema = Temas.objects.get(slug=slug)
if request.method == 'POST':
form = FormNuevoVideo(request.POST)
if form.is_valid():
...
nuevo_video.save()
return redirect('videos:videos_tema', slug=tema.slug, queryset='recientes')
else:
return redirect('videos:nuevo_video', slug=tema.slug) #this same view.
else:
form_nuevo_video = FormNuevoVideo()
context = {'form_nuevo_video': form_nuevo_video, 'tema': tema}
return render(request, template, context)
And in the HTML:
{% block form %}
<form action = "{% url 'videos:nuevo_video' tema.slug %}" method = "post">
{% csrf_token %}
{{form_nuevo_video|crispy}}
<input class = "btn pull-right" type = "submit" value ="enviar"/>
</form>
{% endblock form %}
So, lets say, when someone tries to submit a video with a title of more than 150 characters, it doesn't display the error. I am sure I am missing something simple. Also, I'd like to customize the error messages so that they are in spanish. Thanks in advance.
I have a form on my website, that creates an entry in database. So every time when I refresh a page I got this message first:
The page that you're looking for used information that you entered.
Returning to that page might cause any action you took to be repeated.
Do you want to continue?
Obviously I don't want have the same information more than once in my database.
just in case: this is my code (I know there is a lot of crap that needs to be deleted):
#views.py
#login_required
def subject(request,username, subject_name):
subject_id = Subjects.objects.filter(user = request.user).get(name=subject_name)
#Upload form
if request.method == "POST":
if "upload-b" in request.POST:
form = ContentForm(request.POST, request.FILES, instance=subject_id)
if form.is_valid(): # need to add some clean functions
up_f = FileDescription.objects.get_or_create(subject=subject_id,
subject_name=subject_name,
file_type=request.POST['file_type'],
file_uploaded_by = username,
file_name=request.POST['file_name'],
file_description=request.POST['file_description'],
image = request.FILES['image'],
)
form = ContentForm()
#Show uploaded files with respect to clicked session (Homework, Class , Random ... )
homework_files = Homework.homework.filter(subject_name__exact=subject_name,
file_uploaded_by__exact=username)
class_files = ClassPapers.classpapers.filter(subject_name__exact=subject_name)
random_files = RandomPapers.randompapers.filter(subject_name__exact=subject_name,
file_uploaded_by__exact=username)
return render_to_response('subject_content.html', {'form':form,
'subject_name': subject_name,
'class_files': class_files,
'homework_files': homework_files,
'class_files': class_files,
'random_files': random_files,
},
context_instance=RequestContext(request))
#forms.py:
class ContentForm(forms.ModelForm):
file_name =forms.CharField(max_length=255, widget=forms.TextInput(attrs={'size':20}))
file_description = forms.CharField(widget=forms.Textarea(attrs={'rows':4, 'cols':25}))
class Meta:
model = FileDescription
exclude = ('subject', 'subject_name', 'file_uploaded_by')
#template
<div id="sbj-creation-frm">
<h3>Upload File</h3>
<form action="." method="post" enctype="multipart/form-data">{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="submit" name="upload-b" class="btn-create" />
</form>
</div>
This message is from the browser; and it will display anytime you try to refresh a page that was displayed as the result of a POST request.
It has no bearing on your code, the browser will display the same message on all websites where you try to refresh the page (hit F5 for example) which was displayed as a result of a previous POST request.
To prevent this from happening, make sure all POST requests redirect to a different view upon completion; and not render templates themselves.
redirect to same page working for me :
header("Location: #");
Just redirect your page to current page after inserting
, it will clear all the values and avoid adding the Duplicate records !
example:
protected void btnAdd_Click(object sender, EventArgs e)
{
//your code
Response.Redirect("Currentpage.aspx",true);
//or
Response.Redirect(Request.Url.AbsoluteUri);
}
I'm using the code found here (SO.com) to use the same template to both add and edit a record, but when I add a new record and click Submit, I get a 404 on the URL http://192.168.1.3:5678/app/student/edit/None/, and I'm not exactly sure why.
Here is the relevant portion of my urls.py:
url(r'^app/lesson/new/$', 'edit_lesson', {}, 'lesson_new'),
url(r'^app/lesson/edit/(?P<id>\d+)/$', 'edit_lesson', {}, 'lesson_edit'),
Here is the relevant portion of my views.py:
def edit_lesson(request, id=None, template_name='lesson_edit_template.html'):
if id:
t = "Edit"
lesson = get_object_or_404(Lesson, pk=id)
stu = get_object_or_404(Student, pk=sid)
if stu.teacher != request.user:
raise HttpResponseForbidden()
else:
t = "Add"
lesson = Lesson()
if request.POST:
form = LessonForm(request.POST, instance=lesson)
if form.is_valid():
form.save()
# If the save was successful, redirect to another page
return view_lessons(request)
else:
form = LessonForm(instance=lesson)
return render_to_response(template_name, {
'form': form,
't': t,
'lesson': lesson,
}, context_instance=RequestContext(request))
And finally, here is my template:
<h1>{{ t }} Lesson</h1>
<form action="/app/lesson/edit/{{ lesson.id }}/" method="post"> {% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit" />
</form>
I'm certain that I'm missing something really easy, but I can't seem to put my finger on it. I'm using Django 1.3.1 if that makes any difference.
Thanks,
MC
There's no need to specify any URL in the form's action attribute. Just do
<form action="" method="post">
and it will POST back to the URL that you originally used to access it, which is what you want.
In add case {{ lesson.id }} is None, because lesson is unsaved Lesson() instance, without pk, so your form is being fired to nonexistent URL.
I recommend separating create and edit views and processing them in different ways (or even inherit generic views - with new class-based generic views it's easy and pleasant).
Also, use {% url %} template tag everywhere instead of hard-coded urls.