I have a view that is an input form that people input their information in (name, address, that sort of thing). They will then click an "ok" button.
After people click "ok" I want them to be redirected to a page that has a table with their inputted information.
Any ideas on how to do this?
So far, I have the first view. When clicking OK all the information is stored in a database. I just don't know how to use it from there.
Note: ModelForms are used
If you look carefully the documentation : https://docs.djangoproject.com/en/dev/topics/forms/#processing-the-data-from-a-form
You could understand that, once you have treated (is_valid() - cleaned_data['']), you have to do a call to
render_to_response('template',{ 'name':name, 'last_name':last_name..}, context_instance=RequestContext(request))
And, in the template called, just make the data you need be shown, as usual..
you could try to pass your newly created object to the next view (which shows the new object) like this, the called view just has to accept your object as parameter.
# ... your form processing here
if form.is_valid():
# create and save your object code here
your_object.save()
return redirect('show_new_data_view', your_object=your_object)
Hope this helps.
Related
I make a project which generates a random book when the user press a button. The view that does that is:
def random_book(request):
cartile=Books.objects.all()
random_item=random.choice(cartile)
return render(request, 'carti/book.html', context={"random_item": random_item})
On the page were you are redirected to see the generated random book I want to add a button that says "add to read later".
I know how to make a manyotmany relationship between profile model and books model, but I have no idea how to write a view that gets the generated random book, adds it to the profile view_later field in the database, delete it with another button, what html tag and what should I write inside for "add to read later" and the delete button.
Some help would be appreciated!
I guess that depends on the flow of you website.
If I interpret your idea correctly, one simple way would be to have a view that takes the id of the selected book and add it to read later collection (in the case of the add).
Doing a lot of suppositions, something like this should do the basic,
urlpatterns = [
...
path('site/read_later_add<int:book_pk>/', views.read_later_add),
]
# View (in site/views.py)
def read_later_add(request, book_pk=1):
# get user profile
profile = Profile.objects.get(user=request.user)
# get the book
book = Book.objects.get(pk=book_pk)
# add book to view_later collection
profile.view_later.add(book) # <- this is it
# give some success message
return render(request, 'read_later_add_success.html')
The remove case would be similar just using profile.view_later.remove(book).
BTW, I did not add any error handling or checks, you should definitely should dot that.
If i understand your problem correctly. You can get the random book by implementing a function that return a random number between 1 and the size of Books column. And pass the return integer of this random function to Books.objects.get(pk=random_int)
I have a simple form which adds personal information of a family. sometime it saves two instance of a person just by one submit. Maybe my mouse has problem and and double clicks instead of one click (it has some problems). I think this is not possible and django accepts just one post request from an instance of a form and not more (maybe it accepts). what if may code has problem? if it is problem of my code, why it happens once a while?
house = get_object_or_404(House, id=code)
if request.method == 'POST':
form = ParentForm(request.POST)
if form.is_valid():
# save it if it's valid
parent = form.save(commit=False)
if parent.living == 0:
parent.in_family = 0
if not parent.guardian:
parent.save()
if parent.guardian and parent.in_family:
parent.save()
I use Django 1.8
Edit to clear: this is not the only view sometime saves twice. Maybe it is a bug in django
To solve this problem, first you need to create unique constraint on the corresponding database table. The real solution is based on the database schema. I don't know what fields (columns) you have in the parent table, you can start from add unique constraint to these two fields: child_id and parent_name.
The other problem is you need to prevent the second click. So basically you need to write some JavaScript code: it listens to the onClick event of the submit button. Once the button get clicked, the listener sets the disabled attribute to that button to prevent further clicks.
I think what I'm trying to achieve is not hard, but I have no clue how to do it hehehehe !
Basically what I need is the feature that we have in Django Admin, when you are creating a new object, if you have a Foreign Key, you can add new data (opening a pop-up), save it and then the select box updates automatically.
What I have is this form:
I know that would be easy to do it with some Javascript, but my point is, Django has some rules, and as far I know, I can't add new data to a form already created, right? Otherwise Django won't validate this form. How could I achieve this?
PS: "Local" is the select box where I want to add new data. The user should be able to create a new Local on this page, instead of going to another page to do it. Thanks :)
Here your question:
I can't add new data to a form already created, right? Otherwise Django won't validate this form. How could I achieve this?
Then the answer:
you are right, django will check values match form value rules. But:
realize that your main form is invoked for twice: on GET and on POST. Between both form executions you make changes on database values trhough your new form. That means that in second main form invocation the value added to database is available:
field1 = forms.ModelChoiceField(queryset= ***1*** )
***1***: on second invocation new value is already available on field1.
Then, you don't should to be afraid about this subject, the new value will be available on form on your main form POST request.
Nothing wrong with updating the value using javascript as long the key in your new combo box has the right key in the database then it should be ok.
Call this function after you saved the last entry.
function refreshLocal(){
$.get(window.location.href, '', function(html){
// change the id to the local combox's id
var serverLocalDropBox = $(html).find('#id_local');
if (serverLocalDropBox.length){
$('#id_local').replaceWith(serverLocalDropBox);
}
})
}
If you don't want to use javascript solution, you can post the form with refresh flag and on the server side if you see that flag just don't validate and return the form as is. Since you have a new entry in the foreignkey it will automatically update the queryset to include the new entry.
function serverRefreshLocal(){
var $form = $('#your_form_id');
$form.append('<input type="hidden" name="refresh" value="true" />');
// you can use ajax submit and ajax refresh here if you don't want to leave the page
$form.submit();
}
// Server Side
def your_form_post_view(request):
if request.POST.get('refresh', 'false') == 'true':
# initial is the trick to save user input
your_form = YourForm(initial=request.POST)
context = {
'form': your_form,
}
return render(request, 'your_template.html', context)
# your view code goes here
I'm very new to programming and especially to Django but can't work out how to use any previous answers to my advantage....
Apologies if my question is too vague but essentially, I have two different apps, let's call them app A and app B, with data on two different databases but apps contain information on the same individual item.
I want to edit this information on my 'edit details' page while keeping the apps as separate as possible (well AppB can know about functions in AppA but not vice-versa)...I guess what I really want is a signal which works like so:
A 'submit' view within AppA which is called when I submit changes to the data (using text boxes). The data for AppA is then saved..
And a signal then sent to AppB which ideally would update its data, before the HttpResponseRedirect is performed.
Unfortunately I can't really get this to work. My problem is that if I put 'request' into the arguments for save_details, I get errors like "save_details() takes exactly 3 arguments (2 given)"....does anyone know a clever way of getting something like this to work?
My submit function in AppA looks something like this...
def submit(self, request, id):
signal_received.send(sender=self, id=id)
q = get_object_or_404(AppA, pk=id)
q.blah = request.POST.get('wibble from the form')
...
return Http.....
in my AppB signals.py file, I have put.
signal_received = django.dispatch.Signal(providing_args=['id'])
def save_details(sender, uid, **kwargs):
p = AppB.objects.get(id=id)
p.wobble = request.POST.get('wobble from the form')
...
signal.received.connect(save_details)
Obviously the above doesn't mention request in its arguments which seems to be necessary but if I add that, I get problems with the number of arguments.
(I have imported all the right things at the top of each file I think...hence me leaving that off.)
Any point about the above would be appreciated....e.g. does "request" need to be the first argument? It didn't seem to like me using "self" before but I have tried to copy as much as possible the example at the bottom of the documentation (https://docs.djangoproject.com/en/dev/topics/signals/) but the extra functionality I need in the signal receiving function is flumoxing me.
Thanks in advance...
I have written a function having the following signature:
def action_handler(request, model):
This action_handler is used from different views and handles the actions for this views. One example is deleting objects. In this example the user selectes some objects, selects the delete action and then the user is presented a page to check whether he/she wants to really delete the selected objects. This is done by the following code:
context = {
'action_name' : selected_action,
'object_list' : object_list,
}
return render_to_response("crm/object_delete_check.html", context,
context_instance=RequestContext(request))
For the case that something goes wrong I want to redirect the user to the view from where the user called the action.
Thus I want to ask here whether it is possible to get the calling view from the request object or somewhere else from.
If the def "def action_handler(request, model):" is called from the view "contacts(request):" then i want to redirect the user to the view "contacts(request):" .
But the clue is I do not want to hard-code it since the def action_handler is called from different views. Using a simple "return" is also not possible, since I want to recall the view completely.
if goback: #goback being whatever criteria means "something went wrong"
default_back_url = "someurl_in_case_the_meta_is_messed_up"
back = request.META.get('HTTP_REFERER',default_back_url) #yeah they spelled referrer wrong
if back:
return HttpResponseRedirect(back)
else:
return HttpResponseRedirect(default_back_url)
while META can be faked, it's harder to fake than GET query strings.
You can pass previous page url through GET parameter:
/object_delete_check/?previous=/contacts/
(see contrib.auth.decorators.login_required for example)