Each user has certain vacations that he adds. I want to render his vacations and not anyone else's. What should I do?
The view below shows all vacations, but I only want to show the vacations of a particular user.
def all_vacations(request):
vacations = Vacation.objects.all()
return render(request, 'homePage.html', {'vacations': vacations})
You have to do something like this:
Your URLs:
path('users', views.users_view, name='users')
path('vacation/<int:pk>/', views.user_vacations, name='user_vacation')
Create a view:
# You have to create a view to list users
# List your users
def users_view(request):
users = User.objects.all()
return render(request, 'your_template.html', {'users': users})
Create a template to list users:
<!--In your template:-->
{% for user in users %}
<ul>
<li>{{ user}}</li>
<ul>
{% empty %}
<!-- Do something -->
{% endfor %}
Lastly, you return what you want:
def user_vacations(request, id):
vacations = Vacation.objects.get(author__id=id)
return render(request, 'homepage.html', {'vacations': vacations})
Related
I am looking to modify a form from the user and return that different form in django, however I have tried many different ways, all in views.py, including:
Directly modifying it by doing str(form) += "modification"
returning a new form by newform = str(form) + "modification"
creating a different Post in models, but then I realized that wouldn't work because I only want one post
All the above have generated errors such as SyntaxError: can't assign to function call, TypeError: join() argument must be str or bytes, not 'HttpResponseRedirect', AttributeError: 'str' object has no attribute 'save', and another authority error that said I can't modify a form or something like that.
Here is a snippet from views.py:
class PostCreateView(LoginRequiredMixin, CreateView):
model = Post
fields = ['content']
title = ['title'] #
template_name = 'blog/post_new.html'
success_url = '/'
def form_valid(self, form):
#debugging
tempvar = (str(form).split('required id="id_content">'))[1].split('</textarea></td>')[0] #url
r = requests.get(tempvar)
tree = fromstring(r.content)
title = tree.findtext('.//title')
print(title)
form.instance.author = self.request.user
if "http://" in str(form).lower() or "https://" in str(form).lower():
if tempvar.endswith(' '):
return super().form_valid(form)
elif " http" in tempvar:
return super().form_valid(form)
elif ' ' not in tempvar:
return super().form_valid(form)
else:
return None
models.py:
class Post(models.Model):
content = models.TextField(max_length=1000)
title = models.TextField(max_length=500, default='SOME STRING') #
date_posted = models.DateTimeField(default=timezone.now)
author = models.ForeignKey(User, on_delete=models.CASCADE)
likes= models.IntegerField(default=0)
dislikes= models.IntegerField(default=0)
def __str__(self):
return (self.content[:5], self.title[:5]) #
#property
def number_of_comments(self):
return Comment.objects.filter(post_connected=self).count()
And in home.html, where the post (along with the title and content) is supposed to be shown:
<a
style="color: rgba(255, 255, 255, 0.5) !important;"
href="{% url 'post-detail' post.id %}">
<p class="mb-4">
{{ post.content }}
{{ post.title }} #
</p>
</a>
The original template I'm modifying can be found here.
Thank you so much for your help, I will be very glad to take any advice!!
Ps: I'm using Python 3.7.4
Create a forms.py file inside the app you are talking about, it should look like this:
from django import forms
from . import models
class YourFormName(forms.ModelForm):
class Meta:
model = models.your_model_name
fields = ['field1', 'field2' ,...] # Here you write the fields of your model, this fields will appear on the form where user post data
Then you call that form into your views.py so Django can render it into your template, like this:
def your_view(request, *args, **kwargs):
if request.method == 'POST':
form = forms.YourFormName(request.POST, request.FILES)
if form.is_valid():
instance = form.save(commit=False)
instance.user= request.user
instance.save()
return redirect('template.html') # To redirect if the form is valid
else:
form = forms.YourFormName()
return render(request, "template.html", {'form': form}) # The template if the form is not valid
And the last thing to do is create the template.html:
{% extends 'base.html' %}
{% block content %}
<form action="{% url 'the_url_that_renders_this_template' %}" method='POST' enctype="multipart/form-data">
{% csrf_token %}
{{ form }}
<button type="submit">Submit</button>
</form>
{% endblock content %}
If you want to take the data from DB submitted in that form, you do so with a new function in views.py:
def show_items(request, *args, **kwargs):
data = YourModelName.objects.all()
context = {
"data": data
}
return render(request, "show_items.html", context)
Then in show_items.html:
{% extends 'base.html' %}
{% block content %}
{% for item in data %}
{{item.field1}}
{{item.field2}}
...
{{The items you want to show in that template}}
{% enfor %}
{% endblock content %}
That's what you wanted to do? If not, add a further explanation on what actually you want to do
I'm not sure how to filter dropdown based on user id.
Not I want for user id 2.
I want exactly like this for user id 2.
Model
#python_2_unicode_compatible # only if you need to support Python 2
class PredefinedMessage(models.Model):
user = models.ForeignKey(User)
list_name = models.CharField(max_length=50)
list_description = models.CharField(max_length=50)
def __str__(self):
return self.list_name
class PredefinedMessageDetail(models.Model):
predefined_message_detail = models.ForeignKey(PredefinedMessage)
message = models.CharField(max_length=5000)
View
class PredefinedMessageDetailForm(ModelForm):
class Meta:
model = PredefinedMessageDetail
fields = ['predefined_message_detail', 'message']
exclude = ('user',)
def predefined_message_detail_update(request, pk, template_name='predefined-message/predefined_message_detail_form.html'):
if not request.user.is_authenticated():
return redirect('home')
predefined_message_detail = get_object_or_404(PredefinedMessageDetail, pk=pk)
form = PredefinedMessageDetailForm(request.POST or None, instance=predefined_message_detail)
if form.is_valid():
form.save()
return redirect('predefined_message_list')
return render(request, template_name, {'form':form})
html file
{% extends "base.html" %}
{% load i18n %}
{% block content %}
<form method="post">{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit" />
</form>
{% endblock %}
You can do it in view itself using
form = PredefinedMessageDetailForm(request.POST or None, instance=predefined_message_detail)
form.fields["predefined_message_detail"].queryset= PredefinedMessage.objects.filter(user=request.user)
But filtering happens based on request.user so it should be logged in.Consider that also. Hope this helps
c.ingredientsI am creating a Flask sample app for learning purposes and I have a Form called CocktailForm that contains a SelectMultipleField. This form should be used for creating new cocktails objects as well as for updating them:
class CocktailForm(Form):
name = StringField('What is the coktail\'s name?', validators=[Required()])
ingredients = SelectMultipleField('Ingredients',
coerce=int,
)
submit = SubmitField('Submit')
For editing I want to load the same form with the cocktail data. It works for name, and for loading all choices but I also want to select those choices that are ingredients of that cocktail:
#main.route('/edit/<int:id>', methods=['GET', 'POST'])
def edit(id):
try:
c = Cocktail.query.filter(Cocktail.id == id).one()
form = CocktailForm()
form.ingredients.choices = [(i.id, i.name) for i in Ingredient.query.all()]
if form.validate_on_submit():
c.name = form.name.data
cocktail_ingredients = Ingredient.query.filter(Ingredient.id.in_(form.ingredients.data)).all()
c.ingredients.extend(cocktail_ingredients)
db.session.commit()
else:
form.name.data = c.name
form.ingredients.default = [(i.id, i.name) for i in c.ingredients]
return render_template('new.html', form=form)
except Exception, e:
print e
return redirect(url_for('.index'))
I get unexpected results since on the HTML those choices are not displayed but when I submit the form it seems like those choices are always selected even if you select a new ones.
The template is quite simple:
{% extends "base.html" %}
{% import "bootstrap/wtf.html" as wtf %}
{% block title %}Cocktails{% endblock %}
{% block page_content %}
<div class="page-header">
<h1>New Cocktail</h1>
{{ wtf.quick_form(form) }}
</div>
{% endblock %}
This is what you can see in in the browser. I am editing Gintonic cocktail which is composed of gin and tonic. However they are not displayed as selected:
Browser
Thanks in advance, any tip will be appreciated
The line form.ingredients.default = [(i.id, i.name) for i in Ingredient.query.all()] does not set the selected values. You want to change it to be form.ingredients.data = [i.id for i in c.ingredients].
You should have a redirect/render_template in the validate block, but that is optional.
Hi i am trying to create a post form in django where the user creates a post and the post is displayed back to them on the same page so far i have been unsuccessful. My articles display on the page but the form to post articles doesnt only the submit button shows.
views.py
def articles(request):
args = {}
args.update(csrf(request))
args ['posts'] = post.objects.filter(user = request.user)
args ['full_name'] = User.objects.get(username = request.user.username)
return render_to_response('articles.html', args)
def article(request, post_id=1):
return render(request, 'article.html',
{'post': post.objects.get(id=post_id) })
def create(request):
if request.POST:
form = PostForm(request.POST, request.FILES)
if form.is_valid():
a = form.save(commit=False)
a.user = User.objects.get(username = request.user.username)
a.save()
messages.add_message(request, messages.SUCCESS, "You Article was added")
return HttpResponseRedirect('/posts/all')
else:
form = PostForm()
args = {}
args.update(csrf(request))
args['form'] = form
return render_to_response('articles.html', args)
articles.html
<form action="/posts/create/" method="post" enctype="multipart/form-data">{% csrf_token %}
<ul>
{{form.as_ul}}
</ul>
<input type="submit" name="submit" value="Create Article">
</form>
{% if posts.count > 0 %}
{% for post in posts %}
<div>
<h2>{{full_name}}</h2>
<p>{{ post.body|lower|truncatewords:50 }}</p>
<p>{{post.likes}} people liked this article</a></p>
</div>
{% endfor %}
{% else %}
<p>None to show!</p>
{% endif %}
{% endblock %}
urls.py
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^create/$', 'posts.views.articles'),
url(r'^get/(?P<post_id>\d+)/$', 'posts.views.article'),
url(r'^create/$', 'posts.views.create'),
url(r'^like/(?P<post_id>\d+)/$', 'posts.views.like_article'),
url(r'^article/(?P<post_id>\d+)/$', 'posts.views.add_comment'),
)
First off you aren't passing the form to the template in articles(). You need to have something along the lines of:
args['form'] = YourForm()
In your first view function. Also, in your create view you do this:
a.user = User.objects.get(username = request.user.username)
Which does an unnecessary database lookup of the user again, you can just do this to be clearer and faster:
a.user = request.user
I have a model formset that I want to display 10 forms at a time using Django's Paginator, but it can't be done like paginator = Paginator(formset, 10). What's the correct way to do this, if there is a way?
This is a generic example of the solution I found to my problem:
In the forms.py file:
class MyForm(ModelForm):
class Meta:
model = MyModel
fields = ('description',)
In the views.py file:
from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger
FormSet = modelformset_factory(MyModel, form=MyForm, extra=0)
if request.method == 'POST':
formset = FormSet(request.POST, request.FILES)
# Your validation and rest of the 'POST' code
else:
query = MyModel.objects.filter(condition)
paginator = Paginator(query, 10) # Show 10 forms per page
page = request.GET.get('page')
try:
objects = paginator.page(page)
except PageNotAnInteger:
objects = paginator.page(1)
except EmptyPage:
objects = paginator.page(paginator.num_pages)
page_query = query.filter(id__in=[object.id for object in objects])
formset = FormSet(queryset=page_query)
context = {'objects': objects, 'formset': formset}
return render_to_response('template.html', context,
context_instance=RequestContext(request))
You need to create the formset with the objects in the present page, otherwise, when you try to do formset = FormSet(request.POST, request.FILES) in the POST method, Django raises a MultiValueDictKeyError error.
In the template.html file:
{% if objects %}
<form action="" method="post">
{% csrf_token %}
{{ formset.management_form }}
{% for form in formset.forms %}
{{ form.id }}
<!-- Display each form -->
{{ form.as_p }}
{% endfor %}
<input type="submit" value="Save" />
</form>
<div class="pagination">
<span class="step-links">
{% if objects.has_previous %}
Previous
{% endif %}
<span class="current">
Page {{ objects.number }} of {{ objects.paginator.num_pages }}
</span>
{% if objects.has_next %}
next
{% endif %}
</span>
</div>
{% else %}
<p>There are no objects.</p>
{% endif %}
A more elegant solution is to set ordered=True on the Page object so that it can be passed to a ModelFormSet.
Here is an example:
forms_per_page = 10
current_page = 1
ModelFormSet = modelformset_factory(MyModel, form=MyForm)
queryset = MyModel.objects.all()
paginator = Paginator(queryset, forms_per_page)
page_object = paginator.page(current_page)
page_object.ordered = True
form = ModelFormSet(queryset=page_object)
This is more efficient than the accepted answer because avoids the second database query that takes place in the line:
page_query = query.filter(id__in=[object.id for object in objects])
More correct way to use this
...
formset = FormSet(queryset=page_query.object_list)
...
The problem here is that you're using brands (a Page) in a context that's expecting a QuerySet. So, we need that damn QuerySet. You are in right way, but a lot of code.
In source code we have:
class Page(collections.Sequence):
def __init__(self, object_list, number, paginator):
self.object_list = object_list
self.number = number
self.paginator = paginator
...
So, our queryset in self.object_list attribute and just use it!
formset = SomeModelFormSet(queryset=objects.object_list)
Agree with Elrond Supports Monica. Fake attribute is interesting way to resolve the ordering error (Cannot reorder a query once a slice has been taken.)
But it can be fixed in one line also
queryset = queryset.order_by(Entry._meta.pk.name)
This fake ordering is need for avoid error in django.form.modelsBaseModelFormSet(BaseFormSet).get_queryset(): line #640
that make artificial ordering by pk but it impossible after slicing
(LIMIT-ations in SQL )
More detailed example
queryset = Entry.objects.all()
queryset = queryset.order_by(Entry._meta.pk.name)
paginator = Paginator(object_list=queryset, per_page=10)
page_obj = paginator.get_page(request.GET.get('page'))
EntryFormSet = modelformset_factory(Entry, EntryForm, extra=0)
entryformset = EntryFormSet(queryset=page_obj.object_list)