I see that forms without buttons are very popular (like here). How to create a form that will be automatically submit for two different fields in Django, after the user selects the field (example 1) or type in the text and clicks something (it means completes typing) (example 2):
1.) ChoiceField
forms.py
class Search(forms.Form):
field = forms.ChoiceField(choices=MY_CHOICES)
views.py
if request.method == "GET":
form = Search(request.GET)
if form.is_valid():
print('it's work')
template.html
<form method="GET">
{% csrf_token %}
{{ form }}
</form>
2.) CharField
forms.py
class Search(forms.Form):
field = forms.CharField(max_length=10)
* other files like above
You may simply change forms.py:
class Search(forms.Form):
field = forms.ChoiceField(choices=MY_CHOICES,
widget=forms.Select(attrs={'onchange': 'submit();'}))
Nothing else to add, no jquery needed.
See also here.
You can use jquery in your template like this:
$('#search_field').change(function(){
$('#your_form').submit()
});
or when user click on something:
$('#something').click(function(){
$('#your_form').submit()
});
Related
In my Django project, I want to have the institution selection selected from a list, for this I created a model for the institution name and I want the user to enter it as a pop-up window or a list selection for this:
models.py
class Institution(models.Model):
institutionName = models.CharField(max_length=200,null=True,blank=True)
def __str__(self):
return self.institutionName
views.py
def getInstitutionName(request):
context = {'institutionName':Institution.objects.all()}
return HttpResponse(context)
I created it in the form of html, but I'm having trouble with how to integrate the data I bring here with html. In this process, I want to make a form that includes other entries, only the institution entry in this way. My question on this subject is, what action should I take while printing the data I have brought here to the screen.
Django provides the form field ModelChoiceField. When this field of a form is rendered, it will by default generate a <select> with <option> for each instance in a queryset.
You can either transform this using Javascript, or you can write (or look for) your own widget to use with this field, to generate the HTML you desire.
I'd start by writing a view with such a form and no fancy stuff. When you have it working, then start developing it into the pop-up you seek.
class InstitutionSelectForm ( forms.Form):
institution = forms.ModelChoiceField(
queryset = Institution.objects.all(),
)
# other fields
class InstitutionSelectView( FormView):
form_class = InstitutionSelectView
template_name = 'myapp/institution_select.html'
def form_valid( self, form):
institution_instance = form.cleaned_data['institution']
# do whatever is called for with this institution and
# any other form fields
return HttpResponseRedirect( reverse( ...))
To start with, a very basic template for rendering a form:
<form method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit">
</form>
Make it work, inspect the select structure in your browser, and set to work on customizing it.
Hello everyone I have an HTML form as follows:
and after clicking on post i am redirecting it to views.py. can any one tell me how to get the field values of all the fields of the form into views.py.
here's the output
i want the field value in key value pair like shown in above pic i.e. API=hello&Area=hello1 so on...
i know we can do that using this
if html:
<div class="form-group col-md-2">
<label for="param">TAF Parameter</label>
<input type="text" name="inputdata_API" class="form-control" id="inputapi_param" value="API" readonly>
</div>
and view:
def register(request):
api = request.GET.get['inputdata_API']
But in that case i have to write each and every input name in my view
To get form input without individual access, Django provide ModelForm
For resume :
Define a model to store your form informations
class MyModel(models.Model):
api = models.CharField()
# ...
Define a model form linked to the previous model
from django import forms
class MyModelForm(forms.Form):
class Meta:
model = MyModel
fields = ['api', # all others fields you want to display]
In the views.py
from django.http import JsonResponse
from django.shortcuts import redirect
def register(request):
if request.method == 'POST':
# Instanciate the form with posted data
form = MyModelFor(request.POST)
# Check if form is valid
if form.is_valid:
# Create a new MyModel object if the form is valid
form.save() # This is the most benefit line, save you from request.POST['field_name']
# You can eventually return to the same page
return redirect('.')
else: # The form is invalid return a json response
return JsonResponse({"Error": "Form is invalid"}, status=400)
At the end render the form fields in the template like this :
<form action="{% url 'url_to_register' %}" method="post" novalidate>
{% csrf_token %}
{{ form.as_p }}
<button type="submit" class="btn btn-success">Register</button>
</form>
But the downside of this approach is the style of the form on the frontend :
You need to add some Bootstrap class for example in the right place to make it nice to look, it is a a counterpart...
Django form documentation.
Currently when the user introduces a string in an input field and clicks the submit button, this invokes a view that returns through return render(request, 'index.html', context) a context that
it basically contains data that is displayed in a table.
I would like said table to be visible only after submitting the form and not before and that when it is visible it shows the information obtained from the view.
The problem is that if through inline styling I make this table not visible, for example in the following way:
<div class="row" id="searchVariantTable" style="display: none;">
<!-- SOME TABLE HERE-->
</div>
And then I use the onsubmit event for form or onclick for button, it doesn't work. (It works partially, I can see the tbody but thead is missing, so basically I can't display the retrieved data from the database).
Similarly, if I try something like this:
$('document').ready(function() {
$('#searchVariantTable').hide();
$('form').submit(function(e) {
$('#searchVariantTable').show();
e.preventDefault();
});
});
It doesn't work either.
I think the last option, if I'm not mistaken, is AJAX, but I'm not quite sure how to do something like that with Django (it's my first time using Django)
What am I doing wrong? Is there an option that I am missing?
You can try by using if else in django template:
index.html
<form method="post">
<input type="submit">
</form>
{% if allowed == "yes" %}
<!-- your table code -->
{% endif %}
and in views.py
if request.method == "POST":
return render(request, 'index.html',{'allowed':'yes'})
else:
return render(request,'index.html',{'allowed':'no'})
I have a Django project that has a Students model with multiple fields, and I have implemented a ModelChoiceField form which drops down and allows for selecting a particular record in the Students table.
forms.py:
class StudentChoiceField(forms.Form):
students = forms.ModelChoiceField(
queryset=Student.objects.values_list().order_by("last_name"),
empty_label="(select student)",
widget=forms.Select(attrs={"onChange":'refresh()'})
)
def __init__(self, *args, **kwargs):
super(StudentChoiceField, self).__init__(*args, **kwargs)
# without the next line label_from_instance does NOT work
self.fields['students'].queryset = Student.objects.all().order_by("last_name")
self.fields['students'].label_from_instance = lambda obj: "%s %s" % (obj.last_name, obj.first_name)
The label_from_instance method is overridden, so that the drop-down form displays just two model fields (there are eleven total in the model).
When a student is selected, I want to update some textfields in the page to display the remaining fields of the model. Currently, have implemented a javascript function refresh() which is invoked for the onChange event of the StudentChoiceField form.
index.html (all_students_choice is the StudentChoiceField form):
{% extends "base.html" %}
{% block content %}
<body>
<script>
function refresh(){
var id = document.getElementById("id_students").value;
console.log(id);
}
</script>
<div class="container">
<form method=POST action="">
{% csrf_token %}
{{ all_students_choice }}
</form>
</div>
</body>
{% endblock %}
I have confirmed through the browser console that the javascript function is getting called, and printing the value of the ModelChoiceField form. As expected, after selecting an instance from the dropdown menu the value of the form element is the primary key of the table.
I need advice on the best approach to populate the textfields which I will be adding to display the remaining Student model fields (aside from first and last name). Should these be passed as parameters to the javascript function? Is there a best way to approach this problem.
Answering this question with the approach that that was eventually used, in case it would be of assistance to someone else. Decided to render the same template, but with additional element in the context to reference the selected student. In the initial index home page, the selected_student is None:
def index(request):
....
context = {
'students_choice_ln': students_choice_ln,
'students_choice_fn': students_choice_fn,
'selected_student': None
}
return render(request, 'awards/index.html', context)
For the select function, the selected_student is passed in through the context:
def select(request):
if request.method == "GET":
...
student_id = ...
selected_student = Student.objects.get(pk=student_id)
...
context = {
...
'students_choice_ln': students_choice_ln,
'students_choice_fn': students_choice_fn,
'selected_student': selected_student,
...
}
return render(request, 'awards/index.html', context)
The template can then check whether the selected_student variable is available or not, and then accordingly display the additional fields in a separate div.
If there are any experienced web developers / django developers who see problems with this structure, perhaps they can point them out.
I've implemented TinyMCE with the django-tinymce package. However, my submit button which worked fine without TinyMCE now has become rather useless since I can't submit the form, once everything is filled out.
I can use Ctrl + S inside of TinyMCE (I discovered that by accident) and everything will get submitted correctly. Also, I can use the save-button of the TinyMCE "save" plugin to submit.. Do I have to configure the submit button to make it work with TinyMCE?
Template:
{% extends 'medisearch/header.html' %}
{% load crispy_forms_tags %}
{% block header %}
{{ form.media }}
{% endblock %}
{% block content %}
▷⋅⋅⋅⋅⋅⋅⋅<form action="{{ url }}" method="post">
▷⋅⋅⋅⋅⋅⋅⋅ <div class="form-group">
▷⋅⋅⋅⋅⋅⋅⋅ {% csrf_token %}
▷⋅⋅⋅⋅⋅⋅⋅ {{ form|crispy }}
▷⋅⋅⋅⋅⋅⋅⋅ </div>
▷⋅⋅⋅⋅⋅⋅⋅ <input type="submit" class="btn btn-primary" value="Speichern" />
▷⋅⋅⋅⋅⋅⋅⋅</form>
{% endblock %}
views.py
class EntryDetail(DetailView):
model = Mediwiki
slug_field = 'non_proprietary_name'
template_name = 'mediwiki/entry.html'
class MediwikiForm(FormView):
template_name = 'mediwiki/create.html'
form_class = MediwikiForm⋅
success_url = "/" #TODO user get's redirected to page he's created⋅
def form_valid(self, form):
form.save()
return super(MediwikiForm, self).form_valid(form)
class EntryDisplay(View):
def get(self, request, *args, **kwargs):
try:
view = EntryDetail.as_view()
return view(request, *args, **kwargs)
except Http404: # If there's no entry in db:
if check_user_editor(request.user) == True:
view = MediwikiForm.as_view()
return view(request, *args, **kwargs)
else:
pass
def post(self, request, *args, **kwargs):
view = MediwikiForm.as_view()
return view(request, *args, **kwargs)⋅
forms.py
class MediwikiForm(ModelForm):
wiki_page = forms.CharField(widget=TinyMCE(attrs={'cols': 80, 'rows': 30}))
class Meta:
model = Mediwiki⋅
fields = '__all__'
TinyMCE is in urls.py and under INSTALLED_APPS..
I know it's probably too late for you, but it seems that i had the same issue, just now and my solution might help someone in the future.
You are using crispy, which includes the javascript files for the form on it's own.
Therefore the django_tinymce/init_tinymce.js will be referenced twice.
This will break the submittion of your content, since the form is initialized twice.
In order to fix this you may just remove the call of {{ form.media }}.
I had a similar issue and learned that it has to do with the way that TinyMCE deals with text areas. The following init script worked for me:
<script>
tinymce.init({
selector:'.editor',
setup: function (editor) {
editor.on('submit', function (e) {
editor.save();
});
}
});
</script>
#artifex_knowledge answers makes sense and it works.
To build up on it, besides calling editor.save() on submit (or on change), keep in mind that if users don't fill the text area, they won't be able to submit the form, but the this field is required error message won't be displayed.
This is because the text area field (in this case wiki_page) is required by default, so in the html it will be rendered with required. But TinyMCE hides the text area (and replaces it with an iframe :( ), so if you try to submit the form with an empty required, it won't, but the error message will keep hidden.
(A possible solution is to use JS to remove the required attribute and check it in django later).
Just delete required field from textarea element, which is used as editor.
Deleting the 'required' field in the textarea element solved my problem (like Krysits mentioned)
I also had the same issue as yours, and I just removed for instance: "wiki_page" charfield from the subclass of Modelform, and put Tinymce widget in the Meta class.
class MediwikiForm(ModelForm):
class Meta:
model = Mediwiki⋅
fields = '__all__'
widgets = {
'wiki_page': TinyMCE(attrs={'cols': 80, 'rows': 30})
}