Valid django form not saving - django

In my view for editing friends I'm checking if form is valid, and then save the form. But somehow the data are not updated. Why my updated form is not saved ? It is 100% valid, since I've checked it earlier.
My form :
class FriendForm(forms.ModelForm):
first_name = forms.CharField(widget=forms.TextInput(attrs=dict(attrs_dict, maxlength=50)), label="First name")
last_name = forms.CharField(widget=forms.TextInput(attrs=dict(attrs_dict, maxlength=50)), label="Last name")
pid = forms.RegexField(regex=r'^\d{11}', max_length=11 ,widget=forms.TextInput(attrs=dict(attrs_dict, maxlength=50)))
image = forms.ImageField(label="Image", required=False)
street = forms.CharField(widget=forms.TextInput(attrs=dict(attrs_dict, maxlength=50)), label="Street")
number = forms.CharField(widget=forms.TextInput, label="House/flat number")
code = forms.RegexField(regex=r'^\d{2}[-]\d{3}', max_length=6, widget=forms.TextInput(attrs=attrs_dict), label="Postal code")
city = forms.CharField(widget=forms.TextInput, label="City")
The view :
def edit_friend(request, id):
userprofile = UserProfile.objects.get(user=request.user)
friend = get_object_or_404(Friend, id=id)
if friend.friend_of.id!=userprofile.id:
raise Http404
if request.method == 'POST':
form = FriendForm(request.POST, request.FILES, instance=friend)
if form.is_valid():
form.save()
return HttpResponseRedirect(reverse('user_profile',))
else:
form = FriendForm(instance=friend)
return render_to_response('user/data_operations/edit_friend.html', {
'form':form, 'user':request.user,
}, context_instance=RequestContext(request))
Template :
<form method="post" action="." enctype="multipart/form-data">
<table>
{{ form.as_table }}
<tr>
<td> </td>
<td>
<input type="submit" class="submit" name="submit" value="Save" />
</td>
</tr>
</table>
</form>

I'd need to see the full code for your form to really answer this (can you include it?) but here are some initial thoughts:
Is FriendForm a subclass of django.forms.ModelForm? If so, there's no need to override the save method -- especially since all you're doing here is getting the returned new object, saving it (again), and returning it (again) -- additional processing with no additional benefit.
If FriendForm isn't a subclass of ModelForm, how is it bound to database data? What class is it inheriting from?
UPDATE:
ModelForms aren't connected directly to the database -- they are a shortcut for creating HTML forms for interacting with database models -- i.e. classes inheriting from django.models.Model. You do that by creating a Meta class within your ModelForm.
With a ModelForm, you don't need to manually specify the fields (django does that automatically for you) unless you want to override specific behaviors. You do have to tell django which database Model you're like to use. If you've already defined a database Model, use it; if not, try this:
# in models.py
from django import models
class Friend(models.Model):
first_name = models.CharField( "see <http://docs.djangoproject.com/en/dev/ref/models/fields/> to adjust your syntax" )
... your other fields ...
# in forms.py
from django.forms import ModelForm
from my_project.my_app.models import Friend
class FriendForm(ModelForm):
class Meta:
model = Friend
That's it! Now your FriendForm should work properly. See http://docs.djangoproject.com/en/dev/topics/forms/modelforms/ for more information on using ModelForms.
Alternatively, you don't need to use a ModelForm at all. You could add the following save method to your existing FriendForm:
def save(self):
if self.is_valid():
from myproject.myapp.models import Friend
new_friend = Friend(**self.cleaned_data)
new_friend.save()
return new_friend
... the essence here is that you're importing the Friend model, which encapsulates the database storage behaviors. This is basically what the ModelForm creates automatically for you, but not as robust -- so I recommend using ModelForm.

sometimes, if you add decorators like #transaction.commit_on_success during development, it prevents saving the form if there is even a single error.

I think you may be overriding your model fields with those form fields.
check this. Normally when you have a modelform all you need to do is define a meta class and pass the extra settings you need for that particular form (rendered fields, their widgets, their labels, etc).

You should set action attribute for your form tag. And set url name in urls.py.
Like <form action="{% url 'edit_friend' %}" method="post">
urls.py:
(...'/some-edit-url/', views.edit_friend, name='edit_friend'),

Related

How to reuse Django's admin foreign key widget in admin intermediate pages

I haven't been able to find the answer anywhere on Django's documentation. Though, I'm not surprised given the question is a bit too complex to ask to a search engine.
I'm in a situation where I need to be able reassign a ForeignKey field for one or more entries of a model on Django's admin site.
So far, what I tried to do so using a custom action so that I can select the records I'm interested in and modify them all at once. But, then, I need to select the new related object I want their fk to be reassigned to. So, what I thought to do is an intermediate page where I'd display the fk widget I see all around the admin pages:
But it turns out this widget is really not designed to be publicly used. It's not documented and it's heavily complex to use. So far, I lost several hours digging into Django's code trying to figure how to use it.
I feel like I'm trying to do something really really exotic here so, if there's another solution, I'm all hears.
As shahbaz ahmad suggested, you can use ModelAdmin's autocomplete_fields which creates an select with autocompletion.
But if you're stuck with Django's foreign key widget, because, for instance, you have records which look the same and are indistinguishable in autocomplete, there is a solution.
It turns out ModelAdmin has a get_form method that you can use to retrieve the ModelForm used on the admin page. This method accepts a fields kwargs that you can use to select the fields you want to retrieve in the form. Use it like this:
class MyAdmin(ModelAdmin):
# define the admin subpath to your intermediate page
def get_urls(self):
return [
path(
"intermediate_page/",
self.admin_site.admin_view(self.intermediate_page),
name="intermediate_page",
),
*super().get_urls(),
]
def intermediate_page(self, request):
context = {
# The rest of the context from admin
**self.admin_site.each_context(request),
# Retrieve the admin form
"form": self.get_form(
request,
fields=[], # the fields you're interested in
)
}
return render(request, "admin/intermediate_page.html", context)
If your field is read only – which was my case – there's a workaround to an editable field: you can override the get_readonly_fields method which is called by get_form.
This method accepts an obj parameter which usually takes the model of the object being edited or None when creating a new entry. You can hijack this parameter to force get_readonly_fields exclude fields from read only fields:
def get_readonly_fields(self, request, obj=None):
readony_fields = super().get_readonly_fields(request, obj)
if not (
isinstance(obj, dict)
and isinstance(obj.get("exclude_from_readonly_fields"), Collection)
):
return readony_fields
return set(readony_fields) - set(obj["exclude_from_readonly_fields"])
get_form also has this obj parameter which it passes down to get_readonly_fields so you can call it like this:
# the fields you're interested in
include_fields = []
self.get_form(
request,
obj={"exclude_from_readonly_fields": include_fields},
fields=include_fields
)
Override changelist template of YourModelAdmin class to add one more button apart from add button.
#admin.register(YourModel)
class YourModelAdmin(admin.ModelAdmin):
change_list_template = "custom_your_model_change_list.html"
In custom_your_model_change_list.html,
{% extends "admin/change_list.html" %}
{% block object-tools-items %}
<li>
<a class="button" href="{% url 'your_reassign_url_name' %}">Reassign</a>
</li>
{{ block.super }}
{% endblock %}
Mapped a view to 'your_reassign_url_name' to processed your request.
In urls.py,
urlpatterns = [
path('reassign/', YourReassignView, name='your_reassign_url_name'),
]
forms.py,
class ReassignForm(forms.Form):
# your reassign field
reassign_field = forms.ModelChoiceField(queryset='your queryset')
# Select multiple objects
updatable_objects = forms.ModelMultipleChoiceField(queryset='All objects queryset',
widget=forms.CheckboxSelectMultiple)
In views.py, On GET request you render a form with your required fields and on submit your update your data (reassign values) and after that you redirect admin change_list page.
def YourReassignView(request):
if request.method == 'POST':
form = ReassignForm(request.POST)
if form.is_valid():
# you get value after form submission
reassign_field = form.cleaned_data.get('reassign_field')
updatable_objects = form.cleaned_data.get('updatable_objects')
# your update query
YourModel.objects.filter(
id__in=updatable_objects.values('id')
).update(field_name=reassign_field)
#after that you redirect admin change_list page with a success message
messages.success(request, 'Successfully reassign')
return redirect(reverse('admin:app_label_model_name_changelist'))
else:
context_data = {'form': form}
return render(request, 'reassign_template.html', context=context_data)
else:
form = ReassignForm()
context_data = {'form': form}
return render(request, 'reassign_template.html', context=context_data)
In reassign_template.html,
<form method="POST" action="{% url 'your_reassign_url_name' %}">
{% csrf_token %}
{{ form.as_p }}
<br>
<input type="submit" value="submit">
</form>

How do I catch IntegrityError in Django when using UniqueConstraint across multiple fields?

I have a model with user as 1 field (Foreign Key) and one other field skill_group. I need to make sure the user does not add duplicate skill groups so I added a UniqueConstraint. This is working as the system errors out with IntegrityError at /skillgroup/create/
duplicate key value violates unique constraint "unique_skillgroup" - How do I catch this exception and notify user if duplicate; otherwise save it?
New to Django/Python/Postgres and I thought I could handle it by overriding the save() function, but there is no access to user which is part of the check and I have read this should not be handled here. Is there a try/save catch/message I should be employing? I have tried a few things with no luck. I have seen similar questions on here, but they have not helped. Any help is appreciated.
models.py
class SkillGroup(models.Model):
user = models.ForeignKey(User, on_delete=models.CASCADE)
skill_group = models.CharField(max_length=35)
sequence = models.IntegerField(default=999)
class Meta:
constraints = [
models.UniqueConstraint(fields=['user', 'skill_group'], name='unique_skillgroup'),
]
def __str__(self):
return self.skill_group
def get_absolute_url(self):
return reverse('skillgroup-list')
views.py
class SkillGroupCreateView(LoginRequiredMixin, CreateView):
model = SkillGroup
fields = ['skill_group']
def form_valid(self, form):
form.instance.user = self.request.user
form.instance.sequence = SkillGroup.objects.filter(user=self.request.user).order_by('sequence').last().sequence + 1
return super().form_valid(form)
skillgroup_form.html
{% extends "recruiter/baseskills.html" %}
{% load crispy_forms_tags %}
{% block content%}
<div class="content-section">
<form method="post">
{% csrf_token %}
<fieldset class="form-group">
<legend class="border-bottom mb-4">Skill Group</legend>
{{ form|crispy }}
</fieldset>
<div class="form-group">
<button class="btn btn-outline-info" type="submit">Add Skill Group</button>
</div>
</form>
</div>
{% endblock content%}
I want to either catch the exception and save the record if not a duplicate or put message on screen saying "Skill Group already exists" and leave user on create page. Also, I could remove the UniqueConstraint and handle with code if that is the best solution.
You are inadvertently bypassing Django's form validation here and then trying to save invalid input to the database, which is why Django is feeding back an ugly IntegrityError from the database instead of handling the error gracefully.
If you submit a duplicate User and SkillGroup in your form, your CreateView will helpfully return the error message back into your form template:
"Skill group with this User and Skill group already exists."
But it can only do this if you include a User field in your form. I assume you have excluded User to keep the form template tidy, but that prevents Django's form validation from checking if the combination already exists.
To get around this, add User to your form field as a hidden input. I don't think that's possible using CreateView's behind-the-scenes magic, so you'll need to create a SkillGroupForm to handle that.
# forms.py
from django import forms
from .models import SkillGroup
class SkillGroupForm(forms.ModelForm):
class Meta:
model = SkillGroup
fields = ('user', 'skill_group')
widgets = {
'user': forms.HiddenInput,
}
# views.py
from .forms import SkillGroupForm
class SkillGroupCreateView(LoginRequiredMixin, CreateView):
model = SkillGroup
form_class = SkillGroupForm
def get_initial(self):
return {'user': self.request.user}
def form_valid(self, form):
form.instance.sequence = SkillGroup.objects.filter(user=self.request.user).order_by('sequence').last().sequence + 1
return super().form_valid(form)
The get_initial method passes the request.user as an initial value into the hidden form field, so no user input is needed.
Try to validate skill_group field inside your form class. Define clean_skill_group method like in docs.There you can get queryset of all SkillGroup objects related to your User (like here) and then compare skills. But you need to push somehow your User object or user_id (to get then User object) to the form before form.is_valid() will be called (or call one more time form.is_valid() after). Then show form errors in your html template.
class YourForm(forms.ModelForm):
....some fields, Meta....
def clean_skill_group(self):
your_user_object = ....
previously_created_skills = your_user_object.skill_group_set.all()
skill_input = self.cleaned_data["skill_group"]
if skill_input in previously_created_skills:
raise forms.ValidationError(("This skill group is already exist"), code="invalid") # I suppose you are using model form

How to update ImageField/ FileField using djangorestframework?

I want to update a model using djangorestframework. I don't need to update all fields, so I use PATCH. However, in my form I also have an image field (called 'logo'), which is required for my model. When I try to 'patch' the object and I don't choose a new image for that field, drf throws an error ('logo': 'This field is required').
I know that when using django forms, file fields get a special treatment, meaning that if they already have a value, submitting the form with an empty filefield will just keep the old value. Is there any way to do that using djangorestframework serializers?
Some code for better understanding:
# models.py
class Brand(models.Model):
name = models.CharField(_('name'), max_length=250)
logo = models.ImageField(upload_to='brands/')
# serializers.py
class BrandSerializer(serializers.ModelSerializer):
class Meta:
model = Brand
fields = (
'id',
'name',
'logo',
)
# detail.html
<form method="post" enctype="multipart/form-data">
{%csrf_token%}
<input name="name" type="text" maxlength="30" value="{{ brand.name }}"/>
<input name="logo" type="file" accept="image/*"/>
<input name="_method" type="hidden" value="PATCH">
<input type="submit" value="Update"/>
</form>
The best I could come up with for now was to delete the logo entry from my request.DATA before calling the serializer. I am curious if anyone knows a better solution. Thanks.
Try Link hope fully you will get solution.
Or See this cade, hope fully this will work for you.
class ImageSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Brand
fields = ('name', 'logo')
def saveImage(self, imgFileUri):
#parse dataUri and save locally, return local path
return 'somewhereOverTheBlah'
def restore_fields(self, data, files):
reverted_data = {}
if data is not None and not isinstance(data, dict):
self._errors['non_field_errors'] = ['Invalid data']
return None
for field_name, field in self.fields.items():
"""
So it is iterating over the fields to serialize, when we find the file field
do something different (in this case look for the fileUri field, handle it and replace
it inside of the reverted_data dictionary with the intended file field
"""
if(field_name == 'file'):
field_name = 'dataUri'
field = fields.CharField()
try:
# restore using the built in mechanism
field.field_from_native(data, files, field_name, reverted_data)
# take the dataUri, save it to disk and return the Path
value = reverted_data[field_name]
path = self.saveImage(value)
# set the file <Path> property on the model, remove the old dataUri
reverted_data['file'] = path
del reverted_data[field_name]
except ValidationError as err:
self._errors[field_name] = list(err.messages)
else:
field.initialize(parent=self, field_name=field_name)
try:
field.field_from_native(data, files, field_name, reverted_data)
except ValidationError as err:
self._errors[field_name] = list(err.messages)
return reverted_data

Django Forms choice-like field, without choice limitation

I am writing a writing a webapp that is basically just a form, and it has a button that duplicates a field so that multiple items can be entered. I can't use a SelectMultiple field or any of its variations because there is not a set number of choices to choose from. The user should be able to enter whatever they want into the fields and they must be saved in the model and linked to a record through a manytomany field. Here is a jsfiddle link for demonstration.
HTML
<form>
<label>Field 1
<textarea rows="3"></textarea>
</label>
<label>Multiple Values Possible</label>
<div>
<input type="text">
<select>
<option value="1">1</option>
<option value="2">2</option>
</select>
</div>
<button id="add_div">Add</button>
</form>
JS
add_div = document.getElementById("add_div");
add_div.onclick = function () {
var div = this.previousElementSibling;
var new_div = div.cloneNode(true);
this.parentNode.insertBefore(new_div, this);
return false;
}.bind(add_div);
I cannot figure out how to create the form backend for this. There aren't any field classes that can take in a variable amount of data and validate each one against another field.
What I have tried to do is create a MultiWidget/MultiValueField for the textbox/select dropdown pair, and then subclass my MultiValueField in a class closely following django's ModelMultipleChoiceField. I got stuck trying to get the form field to work with templates, allowing me to add all fields back to the rendered page when rendering with a particular form instance (like how when you use the CheckboxSelectMultiple widget, boxes that are checked in a form instance are rendered checked)
Is there any way to do this and have the ModelForm's save method also save the manytomany fields properly? I know I can override the form's save method and do something like in this stackoverflow question, but I would rather have all the save logic handled by the form fields themselves.
Based on looking at your example jsfiddle, it looks like you don't really need a "Choice Field", what you're looking for are Formsets.
In essence, you would have 2 forms on the page, one which is a normal form and would take care of Field 1, and one which is a Formset which deals with all the many-to-many relations for Field 2
Field2FormSet = formset_factory(Field2ToForm)
Make sure you output the management_form which you can clone with your "add" button.
What you are probably looking for is an inline formset, which can only be used if you are using django models (which you hinted at in your question).
Check out this guide: http://lab305.com/news/2012/jul/19/django-inline-formset-underscore/.
For the lazy, here is a quick example that gets you most of the way there. This app will allow you to continuously add Parent model objects to the database, along with any children that are filled out.
app/models.py
from django.db import models
class ParentModel(models.Model):
parent_field = models.CharField(choices=[(1, 1), (2, 2)])
class ChildModel(models.Model):
parent = models.ForeignKey(ParentModel)
child_field = models.IntegerField(choices=[(1, 1), (2, 2)])
app/views.py
from app import models
from django import forms
from django.forms.models import inlineformset_factory
from django.template import RequestContext, Template
from django.http import HttpResponse
class ParentForm(forms.ModelForm):
class Meta:
model = models.ParentModel
ChildFormSet = inlineformset_factory(models.ParentModel, models.ChildModel)
def test_view(request):
if request.method == "POST":
form = ParentForm(request.POST, request.FILES)
formset = ChildFormSet(request.POST, request.FILES, form.instance)
if form.is_valid() and formset.is_valid():
form.save()
formset.save()
else:
pass # Handle validation error
template = Template(
"<form action='<url for view>' method='post'>"
"{% csrf_token %}"
"{{ form.as_p }}"
"<p>{{ formset.as_table }}</p>"
"<input type='submit' value='Submit'/>"
"</form>"
)
context = RequestContext(request, {
"form": ParentForm(),
"formset": ChildFormSet(),
})
return HttpResponse(template.render(context))
What is shown above will only allow you add up to three children (the default number of extra forms the inline form set produces). To add dynamically, you are going to have to add some java script that creates new forms in the form set on the client side. For that, I suggest you look at the guide I posted above since I don't think I can do better job of explaining it.
Thanks to #Kevin and #Thomas for pointing me towards formsets! Here is how I did it:
models.py
from django.db import models
class RelatedField(models.Model):
field1 = models.CharField(max_length=50)
field2 = models.IntegerField(choices=[(x, x) for x in xrange(1, 11)])
class Record(models.Model):
user = models.ForeignKey(User)
field = models.CharField(max_length=20)
relatedA = models.ManyToManyField(RelatedField, related_name='relatedA')
relatedB = models.ManyToManyField(RelatedField, related_name='relatedB')
views.py
def getIndexContext(data):
if data is None:
recordForm = RecordForm()
relatedFormA = RelatedFormSet(queryset=RelatedField.objects.none(), prefix='related-a')
relatedFormB = RelatedFormSet(queryset=RelatedField.objects.none(), prefix='related-b')
else:
recordForm = RecordForm(data)
relatedFormA = RelatedFormSet(data, prefix='related-a')
relatedFormB = RelatedFormSet(data, prefix='related-b')
return {
'form': recordForm,
'relatedA': relatedFormA,
'relatedB': relatedFormB,
'title': 'Index',
}
def index(request):
if request.method == 'GET':
return render(request, 'record/index.html', getIndexContext(None))
else:
context = getIndexContext(request.POST)
form = context['form']
relatedA = context['relatedA']
relatedB = context['relatedB']
if form.is_valid() and relatedA.is_valid() and relatedB.is_valid():
obj = form.save(commit=False)
obj.user_id = request.user
obj.save()
form.save_m2m()
instances = relatedA.save()
obj.relatedA.add(*instances)
instances = relatedB.save()
obj.relatedB.add(*instances)
return HttpResponse('success!')
else:
return render(request, 'record/index.html', context)
And then some javascript that can duplicate the fields in the formsets, and increment the names by one.

Django and users entering data

I am building a webapp which will be used by a company to carry out their daily operations. Things like sending invoices, tracking accounts receivable, tracking inventory (and therefore products). I have several models set up in my various apps to handle the different parts of the web-app. I will also be setting up permissions so that managers can edit more fields than, say, an office assistant.
This brings me to my question. How can I show all fields of a model and have some that can be edited and some that cannot be edited, and still save the model instance?
For example, I have a systems model for tracking systems (we install irrigation systems). The system ID is the primary key, and it is important for the user to see. However, they cannot change that ID since it would mess things up. Now, I have a view for displaying my models via a form using the "form.as_table". This is efficient, but merely spits out all the model fields with input fields filling in the values stored for that model instance. This includes the systemID field which should not be editable.
Because I don't want the user to edit the systemID field, I tried making it just a label within the html form, but django complains. Here's some code:
my model (not all of it, but some of it):
class System(models.Model):
systemID = models.CharField(max_length=10, primary_key=True, verbose_name = 'System ID')
systemOwner = models.ForeignKey (System_Owner)
installDate = models.DateField()
projectManager = models.ForeignKey(Employee, blank=True, null=True)
#more fields....
Then, my view for a specific model instance:
def system_details(request, systemID):
if request.method == 'POST':
sysEdit = System.objects.get(pk=systemID)
form = System_Form(request.POST, instance=sysEdit)
if form.is_valid():
form.save()
return HttpResponseRedirect('/systems/')
else:
sysView = System.objects.get(pk=systemID)
form = System_Form(instance=sysView)
return render_to_response('pages/systems/system_details.html', {'form': form}, context_instance=RequestContext(request))
Now the html page which displays the form:
<form action="" method="POST">
{% csrf_token %}
<table>
{{ form.as_table }}
</table>
<input type="submit" value="Save Changes">
<input type="button" value="Cancel Changes" onclick="window.location.href='/systems/'">
</form>
So, what I am thinking of doing is having two functions for the html. One is a form for displaying only those fields the user can edit, and the other is for just displaying the content of the field (the systemID). Then, in the view, when I want to save the changes the user made, I would do:
sysValues = System.objects.get(pk=SystemID)
form.save(commit = false)
form.pk = sysValues.sysValues.pk (or whatever the code is to assign the sysValues.pk to form.pk)
Is there an easier way to do this or would this be the best?
Thanks
One thing you can do is exclude the field you don't need in your form:
class System_Form(forms.ModelForm):
class Meta:
exclude = ('systemID',)
The other is to use read-only fields: http://docs.djangoproject.com/en/1.3/ref/contrib/admin/#django.contrib.admin.ModelAdmin.readonly_fields as #DTing suggessted
To make a field read only you can set the widget readonly attribute to True.
using your example:
class System_Form(ModelForm):
def __init__(self, *args, **kwargs):
super(System_Form, self).__init__(*args, **kwargs)
self.fields['systemID'].widget.attrs['readonly'] = True
class Meta:
model = System
or exclude the fields using exclude or fields in the class Meta of your form and display it in your template if desired like so:
forms.py
class System_Form(ModelForms):
class Meta:
model = System
exclude = ('systemID',)
views.py
def some_view(request, system_id):
system = System.objects.get(pk=system_id)
if request.method == 'POST':
form = System_Form(request.POST, instance=system)
if form.is_valid():
form.save()
return HttpResponse('Success')
else:
form = System_Form(instance=system)
context = { 'system':system,
'form':form, }
return render_to_response('some_template.html', context,
context_instance=RequestContext(request))
some_template.html
<p>make changes for {{ system }} with ID {{ system.systemID }}</p>
<form method='post'>
{{ form.as_p }}
<input type='submit' value='Submit'>
</form>