how to add ajax to django webpage - python

I am working on a django application. This application hold two forms. In the first form (image_form) the user can upload an image. In the second form (image_description) the user can fill some description about the image. when the image is uploaded in the first form, an image classifier runs on the image tries to fill parts of the image_description form. Once the second form is submitted the item is displayed in a new page (item_list). This is the urlpatterns for the application.
urls.py
urlpatterns = [
path('', views.index, name='home'),
path('accounts/', include('django.contrib.auth.urls')),
path('signup/', views.signup, name='signup'),
path('items/', views.item_list, name='item_list'),
path('items/upload/description/', views.upload_item, name='upload_item'),
path('items/<int:pk>/', views.delete_item, name='delete_item'),
path('items/upload/image_classification/', views.image_classification, name='image_classification'),
]
here item_list is the page where all items are displayed. This page is displayed once the image_description form is submitted.
upload_item page contains both the forms.
image_classification runs when upload image button in the first form is clicked. This happens in the upload_item page.
views.py
def item_list(request):
items = Item.objects.all()
return render(request, 'item_list.html', {'items':items})
def upload_item(request):
if request.method == 'POST':
form_des = ItemForm(request.POST, request.FILES)
if form_des.is_valid():
form_des.save()
return redirect('item_list')
else:
form_des = ItemForm()
return render(request, 'upload_item.html', {'form_des': form_des})
def image_classification(request):
form_des = ItemForm()
if request.method == 'POST':
if 'file' in request.FILES:
handle_uploaded_file(request.FILES['file'], str(request.FILES['file']))
img = np.expand_dims(cv2.resize(cv2.imread(os.path.join('./media/item/img/', str(request.FILES['file']))), (170, 100)), axis=0)
cat_prediction = cat_classifier.predict_classes(img)[0]
pattern_prediction = pat_classifier.predict_classes(img)[0]
form_des.fields['title'].widget.attrs['value'] = cat_prediction
form_des.fields['pattern'].widget.attrs['value'] = pattern_prediction
form_des.fields['user'].widget.attrs['value'] = request.user
form_des.fields['user'].widget.attrs['readonly'] = True
return render(request, 'upload_item.html', {'form_des': form_des})
else:
return redirect('upload_item')
upload_item.html template
<div class="container">
<div class="row justify-content-center">
<div class="col-7">
<center>
<h2>Upload image</h2>
<!-- <div class="row justify-content-center"> -->
<div class="upload-btn-wrapper">
<form action="{% url 'image_classification' %}" method="POST" enctype="multipart/form-data" data-url="image_classification/" class="my_form">
{% csrf_token %}
<input type="file" name="file" id="file" class="inputfile" multiple/>
<label for="file" class="btn btn-outline-dark btn-lg mt-5 select">Choose a file</label>
<input class='btn btn-primary btn-lg btn-block upload_image_button' type="submit" value="Upload image" disabled/>
</form>
</div>
<!-- </div> -->
<center>
<p class='font-weight-bold mt-5 mb-5 text-danger'>Step: 1 of 2</p>
</center>
<div class="separation"></div>
<h2>Item description</h2>
</center>
<div class="card mb-2 mt-3">
<div class="card-body">
<form method="post" enctype="multipart/form-data" action="{% url 'upload_item' %}" id='item_des'>
{% csrf_token %}
{{form_des.title|as_crispy_field}}
{{form_des.pattern|as_crispy_field}}
{{form_des.color|as_crispy_field}}
{{form_des.user|as_crispy_field}}
<button type="submit" class='btn btn-primary btn-lg btn-block save_btn'>Save item</button>
</form>
</div>
</div>
<center>
<p class='font-weight-bold mt-2 mb-5 text-danger'>Step: 2 of 2</p>
</center>
</div>
</div>
</div>
My problem is when the upload image button is clicked in the first form the url changes from items/upload/description/ to items/upload/image_classification/ and the page reloads with the auto filled sections in the second form.
I want to use AJAX to auto-fill the second form without reloading the page but am not sure how to do it. I followed a few tutorials but am not able to achieve this.
Please help me
Thank you
[EDIT1]
Based on Adeel Siddiqui's answer, I made a few changes to views.py
views.py
def image_classification(request):
form_des = ItemForm()
user =str(request.user)
if request.method == 'POST':
if 'file' in request.FILES:
handle_uploaded_file(request.FILES['file'], str(request.FILES['file']))
img = np.expand_dims(cv2.resize(cv2.imread(os.path.join('./media/item/img/', str(request.FILES['file']))), (170, 100)), axis=0)
cat_prediction = cat_classifier.predict_classes(img)[0]
pattern_prediction = pat_classifier.predict_classes(img)[0]
form_des.fields['title'].widget.attrs['value'] = cat_prediction
form_des.fields['pattern'].widget.attrs['value'] = pattern_prediction
form_des.fields['user'].widget.attrs['value'] = request.user
form_des.fields['user'].widget.attrs['readonly'] = True
context = {
'tops_prediction' :tops_prediction,
'pattern_prediction':pattern_prediction,
'user' :user,
}
return HttpResponse(json.dumps(context))
else:
return redirect('upload_item')
How do I access this context from fillImageDescriptionText in jquery and auto-fill the second form in the upload_item page?
[EDIT-2] screenshot of the output after using the updated answer

You could prevent the default event happening when the form is submitted. Suppose your image classification form has the id image-classify-form, then:
$("#image-classify-form").submit(function(event) {
event.preventDefault();
imageClassifyAjax();
}
where your imageClassifyAjax function does this (i previously forgot to create the FormData object on line 2 which was causing the POST request to fail):
function imageClassifyAjax() {
let $form = $("#image-classify-form");
let form_data = new FormData($form[0]);
$.ajax({
url: $form.attr('action'),
type: $form.attr('method'),
data: form_data,
processData: false,
contentType: false,
dataType: 'json',
success: function (data) {
fillImageDescriptionText(data);
},
error: function (xhr) {
console.log("Something went wrong: " + xhr.responseText);
}
});
}
and fillImageDescriptionText uses the json data returned by the view to fill up the form for the image description. So you have to modify what your view does on the POST method. You have to return an HttpResponse instead of doing a render template:
import json
from django.http import HttpResponse
def image_classification(request):
form_des = ItemForm()
if request.method == 'POST':
...
...
return HttpResponse(json.dumps({'title': cat_prediction, 'pattern': pattern_prediction, 'user': request.user.email}))
fillImageDescriptionText receives this json object as input, so you can basically:
function fillImageDescriptionText(data) {
$("#item_des #id_title").val(data.title);
$("#item_des #id_pattern").val(data.pattern);
$("#item_des #id_user").val(data.user);
$("#item_des #id_user").prop('readonly', true);
}
where id_ is the id generated for the elements by django crispy forms or whatever package you are using for the forms. It is mostly prefixed with an "id_". You can check it up by inspecting elements from the browser.

Related

Django form submit without refreshing

thanks in advance. I know this has been asked a few times. But after reading the previous questions, reading and understanding JSON and AJAX forms tutorials, I still can't find a way to not having the website refreshed after submitting a form. I would really appreciate it if any of you with a higher knowledge of JavaScript is able to give a hand.
This is a newsletter at the bottom part of the Home page that just asks for a name and an email and it keeps the information in the database, it works perfect and I just would like to reply with a confirmation message without refreshing, because it replies a message but the user has to go to the bottom of the page again to see it which is not practical at all.
The HTML is
<div id="contact_form">
<div class="Newsletter">
<form id="form" enctype="multipart/form-data" method="POST" action="" style="text-align: left;">
{% csrf_token %}
<div class="fields">
<div class="fields">
<label for="name" id="name_label">Name</label>
<input type="text" name="name" minlength="3" placeholder="e.g. John Smith" id="name" required>
</div>
<div class="fields">
<label for="email" id="email_label">Email</label>
<input type="email" name="email" placeholder="e.g. john#example.com" id="email" required>
</div>
</div>
<div class="submit">
<button type='submit' id="submit" >Subscribe</button>
</div>
{% include 'messages.html' %}
</form>
</div>
</div>
The index view
def index(request):
"""View function for home page of site."""
if request.method == 'POST':
form = NewsUserForm(request.POST)
if form.is_valid():
instance = form.save(commit=False) #we do not want to save just yet
if NewsUsers.objects.filter(email=instance.email).exists():
messages.warning(request, 'Your email already exists in the newsletter database')
else:
instance.save()
messages.success(request, 'Great! Your email has been submitted to our database.')
try:
send_mail('Welcome ', 'Thank you for subscribing to the Newsletter. ', 'user123#gmail.com',[instance.email], fail_silently=False)
except BadHeaderError: #add this
return HttpResponse('Invalid header found.') #add this
form = NewsUserForm()
return render(request, 'index.html', {'form':form})
Most of the tutorials suggest to create another view + an url for that view + ajax code
I tried this one from here (https://pytutorial.com/how-to-submit-a-form-with-django-and-ajax#top) without success, also eliminating "post" in html method but still not working, even the info got from the form appears in the url. Any help will be welcome, thank you very much.
jquery unsuccessful code
$('#form').on('submit', function(e){
e.preventDefault();
$.ajax({
type : "POST",
url: "{% url 'index' %}",
data: {
name : $('name').val(),
email : $('email').val(),
csrfmiddlewaretoken: '{{ csrf_token }}',
dataType: "json",
},
success: function(data){
$('#output').html(data.msg) /* response message */
},
failure: function() {
}
});
});
unsuccessful ajax view function and url
def ajax_posting(request):
if request.is_ajax():
name = request.POST.get('name', None) # getting data from first_name input
email = request.POST.get('email', None) # getting data from last_name input
if name and email: #cheking if first_name and last_name have value
response = {
'msg':'Your form has been submitted successfully' # response message
}
return JsonResponse(response) # return response as JSON
#path('ajax-posting/', views.ajax_posting, name='index'),# ajax-posting / name = that we will use in ajax url

Can't be redirected after form action in Django

I created a form for creating a new item in Django. The app should be redirected to item_new but it's not working. This below error occurs.
Page not found
http://127.0.0.1.8000/item/new
show.html
<form
action="{% url 'item_new' %}"
method="POST"
enctype="multipart/form-data"
>
{% csrf_token %}
<div class="formItem">
<label>Name</label>
<input type="name" name="name" />
</div>
<div class="formBtn">
<input type="submit" class="btn btnPrimary" value="Submit" />
</div>
</form>
urls.py
urlpatterns = [
path('item/<slug:slug>/', views.item_detail, name='item_detail'),
path('item/new/', views.item_new, name='item_new'),
]
views.py
def item_detail(request, slug):
item = get_object_or_404(Item, slug=slug)
return render(request, 'item/show.html', {'item': item})
def item_new(request):
return redirect('home')
In your urls.py file, change the order of lines. Like this:
urlpatterns = [
path('item/new/', views.item_new, name='item_new'),
path('item/<slug:slug>/', views.item_detail, name='item_detail'),
]
The order of urls, does matter. In your current situation, requests made to http://127.0.0.1.8000/item/new will be sent to item_detail view function instead of item_new.

URL Displaying Wrong Form and View in Django

I am attempting to render a form where the only field is a dropdown. I have the form, view, .html, url all set up. But when I access this url, it shows a different form and view and I suppose also a different .html. I am so confused on why this is happening as it had been working fine for quite some time so I obviously changed something.
forms.py
#this is the form for the dropdown
class ManifestDropDown(forms.Form):
References = forms.ModelChoiceField(queryset=Orders.objects.values_list('reference', flat=True).distinct(),
empty_label=None)
manifest_references.html
<!--html for dropdown-->
{% extends 'base.html' %}
{% block body %}
<div class="container">
<form method="POST" action="manifest">
{% csrf_token %}
{{ reference_list }}
<button type="submit" class="btn btn-primary" name="button">Submit</button>
</form>
</div>
{% endblock %}
views.py
#view for dropdown
def manifest_references(request):
if request.method == 'POST':
if form.is_valid():
reference_id = form.cleaned_data.get('References')
form.save()
query_results = Orders.objects.all()
reference_list = ManifestDropDown()
context = {
'query_results': query_results,
'reference_list': reference_list,
}
return render(request, 'manifest_references.html', context)
urls.py
url(r'^manifest_references', manifest_references, name='manifest_references'),
base.html
<!--showing the link to this url-->
...
<a class="dropdown-item" href="{% url 'manifest_references' %}">Edit Manifests</a>
When I access the url above - instead of showing the manifest_references view with the dropdown, it immediately jumps to a different view manifest which is referenced as the action in the manifest_references.html. Please someone help me determine why this is happening.

how can i access a context of function view with a separate page in base page?

i created newsletters app with below code after that i have a URL : subscribe that perfectly work and save my email after that i want to add this ability in base page for example a user don't go to subscribe page and subscribed i want directly have access in base template and subscribe .
i want to know this because i want add login form to base page and have this problem with that .
this my code and this is my template
tnx for help.
views.py
from django.shortcuts import render
def Subscribe(request):
form = SiqnupNewslettersForm(request.POST or None)
if form.is_valid():
instance = form.save(commit=False)
if SignupNewsletters.objects.filter(email=instance.email).exists():
print("this email already taken ")
else :
instance.save()
context = {
'form':form
}
template_name = "subscribe.html"
return render(request,template_name,context)
def Unsubcribe(request):
form = SiqnupNewslettersForm(request.POST or None)
if form.is_valid():
instance = form.save(commit=False)
if SignupNewsletters.objects.filter(email= instance.email).exists():
SiqnupNewslettersForm.objects.filter(email = instance).delete()
else:
print("your email is not here")
context = {
'form' : form
}
template_name = "unsubscribe.html"
return render(request,template_name,context)
subscribe.html
{% block content %}
<div class="container">
<div class="row">
<form method="POST" action="">
{% csrf_token %}
<div class="form-group">
{{ form }}
</div>
<input type='submit' class="btn btn-primary"
value="submit">
</form>
</div>
</div>
{% endblock %}
urls.py
urlpatterns = [
path('subscribe/',views.Subscribe,name="subscribe"),
path('unsubscribe/',views.Unsubcribe,name= "unsubscribe"),
]
and finally what i have to do for add newsletter form in base page ?
base.html

HTML/Django/Jinja/Python : How to post a fixed value back

This is a HTML template that displays all of the proposals in a database (passed through views.py as a list in the dictionary parameter). I then use a jinja for-loop to go through all the proposals in the database and display their attributes.
How can I Post-request the {{ proposal.id }} back to my python code when the "Learn more" button is clicked? I need this to allow me to display the corresponding values in my other html template.
Sorry if this is a basic question, i'm a high school student and extremely new to django! Thanks alot in advance!
{% block body %}
{% for proposal in proposals %}
<div class="jumbotron">
<h2> Proposal : {{ proposal.title }} </h2>
<h4> Status : {{ proposal.status }} </h4>
<h4> Out of --- Votes: </h4>
<div class="progress">
<div class="progress-bar progress-bar-success" style="width: {{ proposal.votes_for }}%">
<span class="sr-only">35% Complete (success)</span>
{{ proposal.votes_for }}% For
</div>
<div class="progress-bar progress-bar-danger" style="width: {{ proposal.votes_against }}%">
<span class="sr-only">10% Complete (danger)</span>
{{ proposal.votes_against }}% Against
</div>
</div>
<p><a class="btn btn-primary btn-lg" href="#" role="button">Learn more</a></p>
</div>
If you just want to go to the Proposal details you should definitely look to a class-based DetailView.
You can make it with AJAX request or you can make it with form. For the both of types you should have a View to catch it.
HTML Form:
In your template you should have:
<form id="formId" method="post" action="{% url 'catch-proposal' %}">
{% csrf_token %}
<input type="hidden" name="proposal_id" value="{{ proposal.id }}"/>
<p><button type="submit" class="btn btn-primary btn-lg">Learn more</a></p>
<!-- <input type="submit" class="btn btn-primary btn-lg" value="Learn more"/> -->
</form>
It will go to your View from urls.py:
url(r'^post/for/proposal/$', catch_proposal, name='catch-proposal'),
# if your view class-based
# url(r'^post/for/proposal/$', CatchProposal.as_view(), name='catch-proposal')
Then in your view you will catch POST data:
def catch_proposal(request):
if request.method == "POST":
print(request.POST) # have a look for your post params
return reverse_lazy('index') # your response, you can make it on your own
AJAX:
Check it! AJAX and Django
Page uses AJAX without any HTML form
A page makes a POST request via AJAX, and the page does not have an HTML form with a csrf_token that would cause the required CSRF cookie to be sent.
Solution: use ensure_csrf_cookie() on the view that sends the page.
In your scripts define:
function sendPost(proposalId) {
$.ajax({
url: '{% url 'catch-proposal' %}', // or just /ajax/catch/proposal/
method : "POST",
data: {
// your data to send key => val
'id': proposalId
},
dataType: 'json', // it can be xml, json, script, html
success: function (result) {
// Do something if your request was successful (code=200)
// All response data stored in result
console.log(result)
},
error : function(xhr,errmsg,err) {
// Error case
console.log(xhr.status + ": " + xhr.responseText);
}
});
}
For your Learn More button:
<p><button class="btn btn-primary btn-lg" role="button" onclick="sendPost({{ proposal.id }})">Learn more</button></p>
And you will catch it in your View:
#ensure_csrf_cookie # Since you sending POST request without form tag
def catch_proposal(request):
response_data = {} # your response
if request.method == 'POST':
# your post request
if 'id' not in request.POST: # check the param from POST
# Provide error message
response_data['error_message'] = "Can't find ID in POST params..."
else:
# Do whatever
proposal_id = int(request.POST.get('id'))
try:
proposal = Proposal.objects.get(id=transport_id)
response_data['success'] = True
except Proposal.DoesNotExist:
response_data['success'] = False
return JsonResponse(response_data)
else:
response_data = {
'error_message': 'Something is going very strange and wrong...'
}
return JsonResponse(response_data)
Adding created View to urls.py:
from .views import catch_proposal # or yourapp.views
....
url(r'^ajax/catch/proposal/$', catch_proposal, name='catch_proposal'),
....

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