I am new to django (1.2.4). I have created some crud with generic views. But How can I show something like "The student was added successfully" when student is created using django's messaging framework?
As of Django 1.6+, using any class-based generic views, you can rely on the successMessageMixin. It's as simple as adding the mixin to your class definition and setting success_message attribute to whatever you want.
As Olivier Verdier mentioned, please remember to display messages in your main template!
a simple example from the docs:
from django.contrib.messages.views import SuccessMessageMixin
from django.views.generic.edit import CreateView
from myapp.models import Author
class AuthorCreate(SuccessMessageMixin, CreateView):
model = Author
success_url = '/success/'
success_message = "%(name)s was created successfully"
a more complex example:
from django.contrib.messages.views import SuccessMessageMixin
from django.views.generic.edit import CreateView
from myapp.models import ComplicatedModel
class ComplicatedCreate(SuccessMessageMixin, CreateView):
model = ComplicatedModel
success_url = '/success/'
success_message = "%(calculated_field)s was created successfully"
def get_success_message(self, cleaned_data):
# cleaned_data is the cleaned data from the form which is used for string formatting
return self.success_message % dict(cleaned_data,
calculated_field=self.object.calculated_field)
As far as I know, there isn't a straightforward way of doing this using traditional generic views. I've always felt that the documentation on generic views was pretty lacking and so never used them.
In theory you could use a decorator by making the assumption that a redirect meant a successful submission.
So you could write something like this (none of this code has been tested):
urls.py:
try:
from functools import wraps
except ImportError:
from django.utils.functional import wraps
from django.http import HttpRedirectResponse
from django.contrib import messages
from django.views.generic import *
def add_message(success_message=None):
def decorator(func):
def inner(request, *args, **kwargs):
resp = func(request, *args, **kwargs)
if isinstance(resp, HttpRedirectResponse):
messages.success(request, message)
return resp
return wraps(func)(inner)
return decorator
student_info_edit = {
'template_name': 'myapp/student/form.html',
'template_object_name': 'student',
'form_class': studentForm,
}
student_info_new = {
'template_name': 'myapp/student/form.html',
'form_class': studentForm,
'post_save_redirect': '/myapp/students/',
}
urlpatterns += patterns('',
url(r'^students/$', list_detail.object_list, { 'queryset': Student.objects.all() }, name="students"),
url(r'^students/(?P<object_id>\d+)/$', add_message("Student record updated successfully")(create_update.update_object), student_info_edit, name="student_detail"),
url(r'^students/new$', add_message("The student was added successfully.")(create_update.create_object), student_info_new, name="student_new"),
)
All that said and coded, Django 1.3 introduced class-based generic views, so if you're interested in moving onto Django 1.3 you should look into those. They may allow more customization, not sure.
In the long run I rarely see the benefit form using generic views, and this goes double for things like add/update.
The functionality that you are asking for is already implemented in Django generic views:
https://github.com/django/django/blob/1.2.X/django/views/generic/create_update.py#L115
You will see the messages by displaying messages in your main template.
Actually I think the documents explain it pretty well for generic/function based views:
https://docs.djangoproject.com/en/2.0/ref/contrib/messages/
It basically passes context to your template with an if statement to display that context or not.
View:
from django.contrib import messages
def home_page(request):
if request.method == 'POST':
messages.success(request, 'Student added successfully')
context = {}
return render(request, 'homepage/index.html', context)
else:
form =yourForm()
return render(request, 'homepage/index.html', form)
And then it will be displayed in your template using the following. Remember to iterate '...because otherwise the message storage will not be cleared for the next request':
{% if messages %}
<ul class="messages">
{% for message in messages %}
<li{% if message.tags %} class="{{ message.tags }}"{% endif %}>{{ message }}</li>
{% endfor %}
</ul>
{% endif %}
As it renders the page again just add an anchor tag to your form and include in your form action i.e.
action="{% url 'home_page:index' %}#subscribe"
If you're using bootstrap add class alert-success
Related
sorry for all the code given but i have tried so hard for a day still ending up with a problem when passing stage_id as a parameter in the form action url
I have a class Rapports that holds a file (i prefer pdf files) and have a 1to1 relationship with an other class from an other application, i'm sending the stage.id from a html page in Stages app and it's working the id is passed and i can pass it to an other html page but when i write the code i need and passe the same url pattern in the action attribute of the template i want to show(rapport/test.html down below) it return NoReverseMatch and i can't figure out why. Is it because i'm trying to upload a file or is it something else? (1st time working with files)
{% block rpt %}
<li class="nav-item">
<a class="nav-link" href="{% url 'rapport:depo' stages.id %}">
déposer le rapport
</a>
</li>
{% endblock rpt %}
rapport.Rapports
from django.db import models
from satges.models import Stages
class Rapports(models.Model):
stage=models.OneToOneField(
Stages,
on_delete=models.CASCADE,
primary_key=True,
)
src=models.FileField(
("rapport"),
upload_to='rapports/',
max_length=100
)
r_soummit=models.DateTimeField(
auto_now=False,
auto_now_add=False,
)
satges.Stages
class Stages(models.Model):
#Stages attrs that are saved so i think no need to show them here
def est_ete(self):
# comment: this fn returns true if it's a summer intership
if(self.deb.month in[6,7,8]):
return True
# end def
def get_absolute_url(self):
return reverse("satges:det_stg", kwargs={"pk": self.pk})
#saving Stages to database with CreateView is working
def __str__(self):
return self.etd + ' ' + self.nature + " - " + self.au
rapport.forms.py
from django import forms
from .models import Rapports
class RapportsForm(forms.ModelForm):
class Meta:
model = Rapports
fields = ['src']
widgets = {
'src': forms.FileInput(),
}
rapport.views
from django.http import HttpResponse
from django.shortcuts import get_object_or_404, render
from django.views.generic import CreateView
from satges.models import Stages
from .models import Rapports
from .forms import RapportsForm
# Create your views here.
def afficher(request, stage_id):
# comment:
#return HttpResponse('stage: %s'% stage_id)
if request.method == 'POST':
form = RapportsForm(request.POST, request.FILES)
else:
form=RapportsForm()
return render(request, 'rapport/test.html', {'form': form})
class RapportCreateView(CreateView):
model = Rapports
form_class=RapportsForm
template_name = "rapport/test.html"
# end def
'''
def afficher_depo(request, stage_id):
form=RapportsForm()
print(stage_id)
return render(request,'rapport/depo.html', {'form': form})
def depo(request, stage_id):
if request.method == 'POST':
form = RapportsForm(request.POST)
stage=get_object_or_404(Stages, pk=stage_id)
src = form.cleaned_data['src']
rpt=Rapports(stage=stage,src=src)
rpt.save()
else:
return HttpResponse('rapport isn\'t saved')
return render(request,'rapport/depo.html', {'form':form})
'''
the project url conf
from django.contrib import admin
from django.urls import include, path
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('satges.urls')),
path('journal/', include('journaux.urls')),
path('rapport/<int:pk>', include('rapport.urls')),
]+ static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
rapport.urls.py
from django.urls import path, re_path
from . import views
app_name = 'rapport'
urlpatterns = [
path('',views.RapportCreateView.as_view(), name='depo'),
]
test.html (this file is just for testing the main file is called depo.html and it extends otthe page and have the same behavior when i remove the url in action the form appears)
<html>
<head>
<title>test</title>
</head>
<body>
<form method="post" enctype="multipart/form-data" action="{% url 'rapport:depo' satge_id %}">
<!--when i remove the url tag the input form appears and i can choose a file but i ofc can't send it without the action-->
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="déposer votre rapport" name="depo">
</form>
</body>
</html>
this should be all the code related to the problem, i have tried googling, youtube, django doc and with the changes i made along the day the errors differed but the one i couldn't get through is NoReverseMatch
error
Error during template rendering
In template /home/ssoya/Documents/pfe/new/pages/templates/rapport/test.html, error at line 6
Reverse for 'depo' with arguments '('',)' not found. 1 pattern(s) tried: ['rapport/(?P<pk>[0-9]+)\\Z']
after the 1st answer my view became like this but no matter i change it still no use
class RapportsCreateView(CreateView):
model = Rapports
form_class=RapportsForm
template_name = "rapport/depo.html"
def get_context_data(self, **kwargs):
context=super(RapportsCreateView, self).get_context_data(**kwargs)
#stage=Stages.objects.get(pk=self.kwargs.get('pk'))
context['stages'] = Stages.objects.filter(pk=self.kwargs.get('pk'))
return context
then i tried something else: I replaced stage.id with 1 then 13 to see what happens. It raised IntegrityError
NOT NULL constraint failed: rapport_rapports.stage_id
The above exception (NOT NULL constraint failed: rapport_rapports.stage_id) was the direct cause of the following exception:
response = get_response(request)
i'm willing to restructure the question once i found a soloution or understand it more
it's because of stage_id as it's not defined so url will not be found as it needs the parameter stage_id, check the view that rendering your template it is not retuning the stage_id variable
add this to your create view
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super().get_context_data(**kwargs)
# context is a dict so
#update it with your stage id
return context
in this function add the stage_id so it get passed to the template
i can say it is 90% Solved
the problem i had in this question was related to the context in my view and after the answer of #mohamed and some googling and tries i made the view ended up working and saving the file to the server and the database and putting the id of the Satges object as the id of rapport as the attribute satge_id then redirects the user to an other page
here are the modifications i made
views.py
class RapportsCreateView(CreateView):
model = Rapports
form_class = RapportsForm
template_name = "rapport/depo.html"
def form_valid(self, form):
self.stage_id = self.kwargs['pk']
stage = Stages.objects.get(id=self.stage_id)
form.instance.stage_id = stage.id
form.instance.r_soummit = timezone.now()
return super().form_valid(form)
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
stage = Stages.objects.get(pk=self.kwargs.get('pk'))
context['stage'] = stage
context['stage_id'] = stage.id
return context
models.Rapports
def get_absolute_url(self):
return reverse("rapport:det_rpt", kwargs={"pk": self.pk})
def __str__(self):
return self.src
I created an app called "jobs", basically I'd like to create new "jobs" from the admin console and be able to post it on the jobs.html page.
I created the model and views but I think there is something wrong with the views that doesn't allow me to print the "jobs" on the html template.
Can you please tell me if the error is in views.py?
jobs/models.py
from django.db import models
# Create your models here.
class post_job(models.Model):
posizione= models.TextField(max_length=20)
descrizione= models.TextField(max_length=20)
requisiti= models.TextField(max_length=20)
def __str__(self):
"""String for representing the MyModelName object (in Admin site etc.)."""
return self.posizione
jobs/admin.py
from django.contrib import admin
from .models import post_job
# Register your models here.
admin.site.register(post_job)
jobs/views.py
from django.shortcuts import render
from .models import post_job
# Create your views here.
def viz_job(request):
posizione = post_job.posizione
print(posizione)
return render(request,'jobs/jobs.html',{'posizione':posizione})
Proper answer:
In your views:
from django.shortcuts import render
from .models import PostJob # proper naming
def viz_job(request):
jobs = PostJob.objects.all()
return render(request,'jobs/jobs.html',{'jobs': jobs})
in your template:
<ul>
{% for job in jobs %}
<li>
<h3>{{ job.posizione }}</h3>
<div>
{{ job.descrizione }}
</div>
</li>
{% endfor %}
</ul>
Note that all this is documented.
NB: if you're only interested in those two fields and don't need any of the model's methods, related objects or whatever, you can optimize the query a bit by using a values queryset that will yield dicts with the selected fields instead of full model instances:
jobs = PostJob.objects.values("posizione", "descrizione")
Everything else remains the same.
You have to know what do you want to return for the template, for example in the views.py :
from django.shortcuts import render
from .models import post_job
# Create your views here.
def viz_job(request):
jobs = []
descriziones = []
posizione = Job.objects.all()
for pos in posizione:
jobs.append(pos.posizione)
descriziones.append(pos.descrizione)
context = {
'posizione': jobs,
'descrizione': descriziones
}
return render(request, 'jobs/jobs.html',
context=context) # this will return context dictonary to the template
You can filter and get to fetch specific data from your database
Level: Absolute Beginner, trying to build an app to perform some db operation through web UI
models.py
from django.db import models
class MysqlUser(models.Model):
username = models.CharField(max_length=100)
password = models.CharField(max_length=50)
environment = models.CharField(max_length=50)
forms.py
from django import forms
from onboard_app.models import MysqlUser
class MysqlUserForm(forms.ModelForm):
CHOICES = (
('test', 'Test'),
('develop', 'Develop'),
('staging', 'Staging'),
)
environment = forms.MultipleChoiceField(choices=CHOICES)
password = forms.CharField(widget=forms.PasswordInput)
class Meta:
model = MysqlUser
fields = ('username', 'password', 'environment')
views.py
from django.shortcuts import render
from onboard_app.serializers import MysqlUserSerializer
from rest_framework import generics
from onboard_app.forms import MysqlUserForm
from onboard_app.models import MysqlUser
from django.views.generic.edit import CreateView, UpdateView, DeleteView
class MysqlCreateView(CreateView):
model = MysqlUser
form_class = MysqlUserForm
template_name = 'mysqluser_form.html'
success_url = '/mysql/user/add/'
mysqluser_form.html
{% extends "myapp/base.html" %}
{% block title %}MyApp{% endblock %}
{% block content %}
<h1>MySQL User Access</h1>
<form method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" class="btn btn-primary" value="Grant Access">
</form>
{% endblock %}
I'm trying to get Value(s) of field or MultipleChoiceFiled environment after the user Form Submit, and loop through the entered values to perform some action. I have been trying this for long, still can't figure out how. I don't want to process anything in the frontend html. I'm thinking it has to be processed in the Views but not sure how to get the values of the field and loop over.
Any examples or any django concepts to look into will help me a lot. Any help is appreciated, thanks in advance!
Unless you have already done so, I recommend going through Django docs on class-based views (https://docs.djangoproject.com/en/3.0/topics/class-based-views/intro/) to get an overview of how the whole thing works.
So after you submit a form, a post() method of your view is called. CreateView provides a default implementation, which validates the user input using the MysqlUserForm you have provided and then creates an instance of MysqlUser and redirects to the success_url.
If you want to add more logic, you need to overwrite the post() method (or some other method, called by post(), in particular, the form_valid method) and put your logic there. To get a complete sense of how things work, I recommend you to read through the CreateView, BaseCreateView, ProcessFormView, and ModelFormMixin source code, although the inheritance looks a bit complicated (and it is). More to that, I really advise you to walk through the request processing from the View.dispatch method all way to the form_valid method using the debugger and see how things really work. Trust me, it will really contribute to your development skills improvement and understanding. Actually, I've discovered the form_valid method to write this answer by reading the source code (I use rest-framework nowadays and don't remember a lot about Django views).
So, what you need is
class MysqlCreateView(CreateView):
model = MysqlUser
form_class = MysqlUserForm
template_name = 'mysqluser_form.html'
success_url = '/mysql/user/add/'
def form_valid(self, form):
environment = form.cleaned_data['environment']
# insert your code here
return super().form_valid(form)
P.S. A good IDE like a PyCharm is really much much more convenient to read source code, jump to relevant parts and debug than any text editor and PDB debugger.
It's my first time to use ListView and it doesn't work and give me error.
I put get_query but they still give me same error. How can I fix the problem?
And everytime when I write code in views.py I always used 'def' not 'class'. But could see many people use (and also django documents) 'class' for ListView. So for general render stuffs we use 'def' and for django.views.generic stuffs we use class right? Why they distinguished these two?
This is error what I got.
ImproperlyConfigured at /search/results
ListView is missing a QuerySet. Define ListView.model, ListView.queryset, or override ListView.get_queryset().
urls.py
from django.urls import path
from django.conf import settings
from django.views.generic import ListView, TemplateView
from . import views
app_name = 'search'
urlpatterns = [
path('', TemplateView.as_view(template_name = "search/index.html")),
path('results', ListView.as_view(template_name = 'search/results.html')),
path('beerlist', views.beerlist, name='beerlist'),
path('<int:beerinfo_id>', views.beerinfo, name='beerinfo'),
]
views.py
from django.shortcuts import render, get_object_or_404, redirect
from django.db.models import Q
from django.views.generic import ListView, TemplateView
from .models import Beerinfo
# Create your views here.
def index(TemplateView):
template_name = 'search/index.html'
def results(ListView):
model = Beerinfo
template_name = 'search/results.html'
def get_queryset(self):
query = self.request.GET.get('q')
object_list = Beerinfo.objects.filter(
Q(name__icontains = query) | Q(label__icontains = query)
)
return obejct_list
index.html
<form action="{% url 'search:results' %}" name="se">
<label for='search'>What do you want to find?</label>
<input type="text" name='q'>
<input type="submit">
</form>
results.html
<ul>
{% for beer in ojbect_list %}
<li>{{ beer.name }}</li>
{% endfor %}
</ul>
models.py
from django.db import models
# Create your models here.
class Beerinfo(models.Model):
name = models.CharField(max_length=100)
label = models.CharField(max_length=500)
def __str__(self):
return self.name
You need to define the class that the list view will work with. For example:
class UserListView(ListView):
model = User
You can use a function (def) to accomplish the same thing that a generic view class, the difference is that most of what you write in the function is already defined in the class. In my example above, that class already handles the rendering of a default template, a context with the list of object of that template and pagination. The idea is to keep your code DRY.
The second advantage is that it creates a standard for your code, for example the default template to be used is
%(app_label)s/%(model_name)s%(template_name_suffix)s.html, so if your app name is users and your model is User, the this view expects a template named users/userlist.html
To use the pagiation simply set the paginate_by attribute of the class.
If you are trying to implement a simple view (for example all CRUD actions, then is very likely that you will benefit from using clases. Another good thing that classes give you, is that you can inherit goodies, for example, you can create a BaseListView class that inherits from ListView and set paginate_by to 25. If all your clases inherit from BaseListView then all your list will be paginated by 25 elements.
In views.py change def to class , you need to define a class to use Listview, Class Results(ListView). In urls.py, you are calling Listview , you should call views.Results.as_view()
I know this question was asked before, but none worked for me. I have this code that I want it to be executed when a button is clicked and a message is passed
import time
from sinchsms import SinchSMS
number = '+yourmobilenumber'
message = 'I love SMS!'
client = SinchSMS(your_app_key, your_app_secret)
print("Sending '%s' to %s" % (message, number))
response = client.send_message(number, message)
message_id = response['messageId']
response = client.check_status(message_id)
while response['status'] != 'Successful':
print(response['status'])
time.sleep(1)
response = client.check_status(message_id)
print(response['status'])
Basically, what I need is to add an input in a template "HTML File", this input get passed to the message variable in the code above, same with the number. I can easily do that with instances, but how can the below get executed when a button is clicked from the form in the template?
I'm kinda newbie in Django and still finding my way
Here is the tutorial that explains how to make the python file, but execute it from the shell, not a django application.
I hope I was clear describing my problem and any help would be appreciated!
All you need is a form with a message field. In a view, you want to show that form and when the user press submit, you want to execute your script.
Here is some pseudo-code:
urls.py
url('^my-page/' my_views.my_view, name='my-page'),
forms.py
SmsForm(forms.Form):
message = fields.CharField(...)
my_views.py
def my_view(request):
form = SmsForm(data=request.POST or None)
if request.method == 'POST':
if form.is_valid():
send_sms(form.cleaned_data['message']) # do this last
messages.success(request, "Success")
return HttpResponseRedirect(request.path)
else:
messages.warning(request, "Failure")
return render(request, 'my_template.html', {'form': form})
Check the Django documentation about urls, views, forms and messages and proceed step by step:
get the page to load
get the form to load
get the form submission to work and simply show "Success" or "Failure"
finally, write the send_sms function (you've almost done it)
Lets start from the dust cloud.
What you are asking is mostly about how the web pages work. You need to know how to pass parameters using HTML. There are lots of ways to do it. But with django there is a pattern.
You need a url, and a view to catch any requests. Then you need to create a template and a form inside it. With this form you could create some requests to send data to your view.
To create you need to edit urls.py inside your project add an url:
urls.py
from django.conf.urls import url
from my_app.views import my_view
urlpatterns = [
...
url(r'^my_url$', my_view, name='my_view')
...
]
For more about urls please look at URL dispatcher page at documentation.
Then create your view inside your app which is my_app in my example. Edit my_app/views.py
my_app/views.py
from django.http import HttpResponse
def my_view(request):
return HttpResponse('IT WORKS!')
This way you get a working view which could be accessed with path /my_url. If you run ./manage.py runserver you could access your view from http://localhost:8000/my_url.
To create a form you need to create a template. By default django searches app directories for templates. Create a templates directory in your app, in our case my_app/templates and create an HTML file inside. For example my_app/templates/my_form.html. But i advice to create one more directory inside templates directory. my_app/templates/my_app/my_form.html. This will prevent template conflicts. You can check Templates page at documentation for more.
my_app/templates/my_app/my_form.html
<html>
<body>
<form action="/my_url" method="POST">
{% csrf_token %}
<input type="text" name="number">
<input type="text" name="message">
<input type="submit" value="Run My Code">
</form>
</body>
</html>
This is the one of the ways of creating your form. But I do not recommend it. I will make it prettier. But first lets "Make it work", edit your views.py:
csrf_token is a django builtin template tag, to put CSRF token into your form. By default django requires CSRF tokens at every post
request.
my_app/views.py
from django.http import HttpResponse
from django.shortcuts import render
def my_view(request):
if request.method == 'GET':
return render('my_app/my_form.html')
elif request.method == 'POST':
# get post parameters or None as default value
number = request.POST.get('number', None)
message = request.POST.get('message', None)
# check if parameters are None or not
if number is None or message is None:
return HttpResponse('Number and Message should be passed')
# your code goes here
...
return HttpResponse('Your code result')
Till this point the purpose of this answer was "Making it work". Lets convert it nice and clean. First of all we would create Form. Forms are like models, which helps you create forms as objects. It also handles form validations. Forms are saved inside forms directory generally. Create my_app/forms.py and edit it:
my_app/forms.py
from django import forms
class MyForm(forms.Form):
number = forms.CharField(max_length=15, required=True)
message = forms.CharField(max_length=160, required=True)
Put your form inside your template:
my_app/templates/my_app/my_form.html
<html>
<body>
<form action="{% url 'my_view' %}" method="POST">
{% csrf_token %}
{{ form }}
</form>
</body>
</html>
Besides the form, the action of the HTML form tag is also changed.
url template tag is used to get url form url name specified in urls.py.
Instead of url tag, {{ request.path }} could have been used.
Create a form instance and pass it to the template rendering:
my_app/views.py
from django.http import HttpResponse
from django.shortcuts import render
from .forms import MyForm
def my_view(request):
if request.method == 'GET':
form = MyForm()
return render('my_app/my_form.html', {'form': form})
elif request.method == 'POST':
form = MyForm(request.POST)
# check if for is not valid
if not form.is_valid():
# return same template with the form
# form will show errors on it.
return render('my_app/my_form.html', {'form': form})
# your code goes here
...
return HttpResponse('Your code result')
You can use class based vies to write your view, but it's not necessary. I hope it helps.
You can create a view that takes up query parameters from the url and use it for further implementation. Then you can create a link/button in the html template which can redirect you to that url. For example:
in urls.py:
url(r'^run_a/(?P<msg>\w{0,25})/(?P<num>\w{0,25})/$', yourcode, name='get_msg'),
in template:
submit
in views.py:
def get_msg(request,msg,num):
message=msg
number=num
#rest of the code
Hope this helps :)