Most of the info I find online is for multiple checkboxes. I just want 1.
I have:
class CategoryForm(FlaskForm):
category = StringField('category',validators=[DataRequired()])
checkbox = BooleanField('Private?')
#app.route('/category/<categoryid>',methods=('GET','POST'))
def category(categoryid):
category = Category.query.get(categoryid)
if request.method == 'POST':
if request.form.get('category'):
category.name = request.form['category']
category.private = request.form['private']
db.session.add(category)
db.session.commit()
return redirect(url_for('index'))
c_form = CategoryForm()
c_form.category.data = category.name
return render_template('category.html',form =c_form,category=category)
And my 'category' template:
<form method="post">
{{ form.hidden_tag() }}
{{ form.checkbox }}
<button type="submit">Go!</button>
</form>
right now my browser renders this:
<peewee.BooleanField object at 0x105122ad0> Go!
Obviously I would like it to render the checkbox instead. How can I do this? Do I need a widget ?
I'm having the impression that you're using the fields from peewee as the fields in your form, that isn't going to work. The most likely case is that you're importing both and one import is overwriting the other.
If you need to have both the model and the form in the same file, use aliases.
from peewee import BooleanField as PeeBool
from wtforms import BooleanField as WTBool
Related
I've have a view that uses the formset factory but I'm trying to style the rendered html form. But the usual styling for forms doesn't work and im not sure what i need to do to apply the CSS stytles.
My View
def get_questionnaire(request, project_id, questionnaire_id):
# Get the Response object for the parameters
response = Response.objects.get(
project_name_id=project_id, questionnaire_id=questionnaire_id
)
AnswerFormSet = modelformset_factory(Answer, fields=('answer',), extra=0)
answer_queryset = Answer.objects.filter(response=response
).order_by('question__sequence'
).select_related('question')
if request.method == 'POST':
# to be completed
pass
else:
# Get the list of questions for which no Answer exists
new_answers = Question.objects.filter(
questionnaire__response=response
).exclude(
answer__response=response
)
# This is safe to execute every time. If all answers exist, nothing happens
for new_answer in new_answers:
Answer(question=new_answer, response=response).save()
answer_formset = AnswerFormSet(queryset=answer_queryset)
return render(request, 'pages/formset.html', {'formset': answer_formset})
for the moment im just trying to style a single field by applying the widgets in forms.py. But this isn't working.
class answer_formset(ModelForm):
class Meta:
model = Answer
fields = ('answer',)
widgets = {
'answer': forms.Select(attrs={'class': 'form-select'}),
}
HTML
{{ formset.management_form }}
{% for form in formset %}
{{ form.id }} {{ form.instance.question.question }}
{{ form.answer }}
<br>
{% endfor %}
Thanks
You need to tell the factory which ModelForm to use:
AnswerFormSet = modelformset_factory(Answer, form=answer_formset, fields=('answer',), extra=0)
From docs:
You can optionally pass a form argument to use as a starting point for
constructing the ModelForm.
more info: https://docs.djangoproject.com/en/4.0/ref/forms/models/#modelform-factory
I want to use more than one forms in the same page from the same model.
Ok, lets take it easy.
I have Social modules that have 3 attributes (network, url and image) charfield.
I've added 4 values in Social database (Facebbok, Twitter, Youtube, Pinterest)
In the settings view (settings.html) i want to have all 4 forms (for Facebook, Twitter etc.) to edit them.
Something like that:
Facebook: text input (that contains the current facebook url)
Youtube: text input (that contains the current youtube url)
etc.
So when i go to settings.html i can change, update social url for all networks.
I have something like this for General Settings module (that have 3 fields, Title, Slug, Description and 1 attribute cuz the website have 1 title, 1 slug and 1 description). For this one is pretty simple i can use get_object_or_404 because Settings module have just 1 value and i can select it.. but the problem is Social Module have more values and i want to have on my page all forms from them so i can edit how ever i want.
views.py
def settings(request):
sidebar_items = Sidebar.objects.order_by('position')
social_items = Social.objects.order_by('network')
settings = get_object_or_404(Settings, pk = 1)
if request.method == "POST":
form_settings = SettingsForm(request.POST, instance = settings)
if form_settings.is_valid():
settings = form_settings.save(commit = False)
settings.save()
return HttpResponseRedirect('/dashboard/settings')
else:
form_settings = SettingsForm(instance = settings)
context = {'sidebar_items' : sidebar_items, 'form_settings' : form_settings, 'social_items' : social_items}
return render(request, 'dashboard/settings.html', context)
Django doesn't care how many forms you want to initialize in your view. If they're for the same model, you can use a formset. Otherwise, you can initialize and create objects however you want.
Example:
def your_view(request):
social_items = Social.objects.order_by('network')
forms = []
for index, social_item in enumerate(social_items):
forms.append(SocialForm(request.POST or None, instance=social_item,
prefix="form_{}".format(index)))
if request.method == 'POST':
for form in forms:
if form.is_valid():
form.save()
# do whatever next
return render(request, 'some-template.html', {'forms': forms})
You don't need three separate form tags in your template. You can submit all of the data in one post. Django will try to hydrate an instance of each model from the POST data, and return any errors if that fails.
In your template, you'll need to iterate over the form instances:
# some-template.html
<form action="." method="post" enctype="x-www-form-urlencoded">
{% for form in forms %}
<ol>
<li>
{{ form.network }}
{{ form.network.errors }}
</li>
<li>
{{ form.url }}
{{ form.url.errors }}
</li>
<li>
{{ form.image }}
{{ form.image.errors }}
</li>
</ol>
{% endfor %}
<button type="submit">Save</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.
Is it possible to populate a password field in wtforms in flask?
I've tried this:
capform = RECAPTCHA_Form()
capform.username.data = username
capform.password.data = password
The form is defined like:
class RECAPTCHA_Form(Form):
username = TextField('username', validators=[DataRequired()])
password = PasswordField('password', validators=[DataRequired()])
remember_me = BooleanField('Remember me.')
recaptcha = RecaptchaField()
The template looks like this:
<form method="POST" action="">
{{ form.hidden_tag() }}
{{ form.username(size=20) }}
{{ form.password(size=20) }}
{% for error in form.recaptcha.errors %}
<p>{{ error }}</p>
{% endfor %}
{{ form.recaptcha }}
<input type="submit" value="Go">
</form>
I have tried to change the PasswordField to a TextField, and then it works.
Is there some special limitation to populating PasswordFields in wtforms?
Update: After looking through the WTForms docs I found an even better solution. There is a widget arg.
from wtforms import StringField
from wtforms.widgets import PasswordInput
class MyForm(Form):
# ...
password = StringField('Password', widget=PasswordInput(hide_value=False))
As yuji-tomita-tomita pointed out, the PasswordInput class (source) has an hide_value argument, however the constructor of PasswordField (source) does not forward it to the PasswordInput. Here is a PasswordField class that initializes PasswordInput with hide_value=False:
from wtforms import widgets
from wtforms.fields.core import StringField
class PasswordField(StringField):
"""
Original source: https://github.com/wtforms/wtforms/blob/2.0.2/wtforms/fields/simple.py#L35-L42
A StringField, except renders an ``<input type="password">``.
Also, whatever value is accepted by this field is not rendered back
to the browser like normal fields.
"""
widget = widgets.PasswordInput(hide_value=False)
Something I've found with Flask, and Flask apps in general, is that the source is the documentation. Indeed, it looks like by default you cannot populate the field. You can pass an argument hide_value to prevent this behavior.
This is a good call, since if you can populate the field, you have access to the raw password... which could be dangerous.
class PasswordInput(Input):
"""
Render a password input.
For security purposes, this field will not reproduce the value on a form
submit by default. To have the value filled in, set `hide_value` to
`False`.
"""
input_type = 'password'
def __init__(self, hide_value=True):
self.hide_value = hide_value
def __call__(self, field, **kwargs):
if self.hide_value:
kwargs['value'] = ''
return super(
I believe there is an easier way to access the data of the password field, without usinghide_value. In your view, simply add in the request data as an argument to the form's constructor:
from flask import request
capform = RECAPTCHA_Form(request.form)
capform.username.data = username
capform.password.data = password
This should make the password input available for validation, and to be used in testing if desired.
I'm new to Django and I'm creating an app to create and display employee data for my company.
Currently the model, new employee form, employee table display, login/logout, all works. I am working on editing the current listings.
I have hover on row links to pass the pk (employeeid) over the url and the form is populating correctly- except the manytomanyfields are not populating, and the pk is incrementing, resulting in a duplicate entry (other than any data changes made).
I will only put in sample of the code because the model/form has 35 total fields which makes for very long code the way i did the form fields manually (to achieve a prettier format).
#view.py #SEE EDIT BELOW FOR CORRECT METHOD
#login_required
def employee_details(request, empid): #empid passed through URL/link
obj_list = Employee.objects.all()
e = Employee.objects.filter(pk=int(empid)).values()[0]
form = EmployeeForm(e)
context_instance=RequestContext(request) #I seem to always need this for {%extend "base.html" %} to work correctly
return render_to_response('employee_create.html', locals(), context_instance,)
#URLconf
(r'^employee/(?P<empid>\d+)/$', employee_details),
# snippets of employee_create.html. The same template used for create and update/edit, may be a source of problems, they do have different views- just render to same template to stay DRY, but could add an additional layer of extend for differences needed between the new and edit requests EDIT: added a 3rd layer of templates to solve this "problem". not shown in code here- easy enough to add another child template
{% extends "base.html" %}
{% block title %}New Entry{% endblock %}
{% block content %}
<div id="employeeform">
{% if form.errors %}
<p style="color: red;">
Please correct the error{{ form.errors|pluralize }} below.
</p>
{% endif %}
<form action="/newemp/" method="post" class="employeeform">{% csrf_token %} #SEE EDIT
<div class="left_field">
{{ form.employeeid.value }}
{{ form.currentemployee.errors }}
<label for="currentemployee" >Current Employee?</label>
{{ form.currentemployee }}<br/><br/>
{{ form.employer.errors }}
<label for="employer" class="fixedwidth">Employer:</label>
{{ form.employer }}<br/>
{{ form.last_name.errors }}
<label for="last_name" class="fixedwidth">Last Name:</label>
{{ form.last_name }}<br/>
{{ form.facility.errors }} #ManyToMany
<label for="facility" class="fixedwidth">Facility:</label>
{{ form.facility }}<br/><br/>
</div>
<div id="submit"><br/>
<input type="submit" value="Submit">
</div>
</form>
#models.py
class Employee(models.Model):
employeeid = models.AutoField(primary_key=True, verbose_name='Employee ID #')
currentemployee = models.BooleanField(null=False, blank=True, verbose_name='Current Employee?')
employer = models.CharField(max_length=30)
last_name = models.CharField(max_length=30)
facility = models.ForeignKey(Facility, null=True, blank=True)
base.html just has a header on top, a menu on the left and a big empty div where the forms, employee tables, etc all extend into.
screenshot2
So, how do I need to change my view and/or the in the template to update an entry, rather than creating a new one? (
And how do I populate the correct foriegnkeys? (the drop down boxes have the right options available, but the "-----" is selected even though the original database entry contains the right information.
Let me know if i need to include some more files/code
I have more pics too but i cant link more or insert them as a new user :< I'll just have to contribute and help out other people! :D
EDIT:
I've been working on this more and haven't gotten too far. I still can't get the drop-down fields to select the values saved in the database (SQLite3).
But the main issue I'm trying to figure out is how to save as an update, rather than a new entry. save(force_update=True) is not working with the default ModelForm save parameters.
views.py
def employee_details(request, empid):
context_instance=RequestContext(request)
obj_list = Employee.objects.all()
if request.method == 'POST':
e = Employee.objects.get(pk=int(empid))
form = EmployeeForm(request.POST, instance=e)
if form.is_valid():
form.save()
return HttpResponseRedirect('/emp_submited/')
else:
e = Employee.objects.get(pk=int(empid))
form = EmployeeForm(instance=e)
return render_to_response('employee_details.html', {'form': form}, context_instance,)
also changed template form action to "" (from /newemp/ which was the correct location for my new employee tempalte, but not the update.
Thanks to this similar question.
updating a form in djnago is simple:
steps:
1. extract the previous data of the form and populate the edit form with this these details to show to user
2. get the new data from the edit form and store it into the database
step1:
getting the previous data
views.py
def edit_user_post(request, topic_id):
if request.method == 'POST':
form = UserPostForm(request.POST)
if form.is_valid():
#let user here be foreign key for the PostTopicModel
user = User.objects.get(username = request.user.username)
#now set the user for the form like: user = user
#get the other form values and post them
#eg:topic_heading = form.cleaned_data('topic_heading')
#save the details into db
#redirect
else:
#get the current post details
post_details = UserPostModel.objcets.get(id = topic_id)
data = {'topic_heading':topic.topic_heading,'topic_detail':topic.topic_detail,'topic_link':topic.topic_link,'tags':topic.tags}
#populate the edit form with previous details:
form = UserPostForm(initial = data)
return render(request,'link_to_template',{'form':form})