Django Forms - Processing GET Requests - validation

We have an existing Django form that accepts GET requests to allow users to bookmark their resulting query parameters. The form contains many fields, most of which are required. The form uses semi-standard boilerplate for handling the request, substituting GET for POST:
if request.method == 'GET':
form = myForm(request.GET)
if form.isValid()
# Gather fields together into query.
else
form = myForm()
The problem is that the first time the form is loaded, there's nothing in the GET request to populate the required fields with, so most of the form lights up with 'missing field' errors.
Setting initial values doesn't work; apparently, the non-existent values in the GET request override them.
How can we avoid this? I'm pretty certain we're simply not processing things correctly, but I can't find an example of a form that handles GET requests. We want errors to show up if the user hits the "Submit" button while fields are blank or otherwise invalid, but don't want these errors showing up when the form is initially displayed.

The positional argument to the forms.Form subclass informs Django that you intend to process a form rather than just display a blank/default form. Your if request.method == 'GET' isn't making the distinction that you want because regular old web requests by typing a URL in a web browser or clicking a link are also GET requests, so request.method is equal to GET either way.
You need some differentiating mechanism such that you can tell the difference between a form display and a form process.
Ideas:
If your processing is done via. AJAX, you could use if request.is_ajax() as your conditional.
Alternatively, you could include a GET token that signifies that the request is processing. Under this example, first you'd need something in your form:
<input type="hidden" name="action" value="process_form" />
And then you can look for that value in your view:
if 'action' in request.GET and request.GET['action'] == 'process_form':
form = myForm(request.GET)
if form.is_valid():
# form processing code
else:
form = myForm()
I'll also give you the standard, boilerplate point that it's generally preferable not to use GET for form processing if you can help it (precisely because you run into difficulties like this since you're using an anomalous pattern), but if you have a use case where you really need it, then you really need it. You know your needs better than I do. :-)

If your clean page load doesn't have any non form GET params, you can differentiate between a clean page load and a form submit in your view. Instead of the usual
form = YourForm()
if request.POST:
you can do
if request.GET.items():
form = YourForm(request.GET)
if form.is_valid():
...
else:
form = YourForm()
If your clean page load could have other params (eg email link tracking params) you'll need to use the QueryDict methods to test if any of your form params are in the request.

request.GET is and empty dictionary when you first load a clean form. Once you have submitted the form, request.GET will be populated with your fields data, even if the fields contain only empty data.

My first question is this, which I posted as comment:
Why not just use request.POST and the standard way of processing form data?
After considering everything here, perhaps what you are looking for is a way of processing data in your query string to populate a form. You can do that without using request.GET as your form.data.
In my own views, I take advantage of a utility function I created to add initial data to the form from request.GET, but I am not going to share that function here. Here's the signature, though. initial_dict is typically request.GET. model_forms is either a single ModelForm or a list of ModelForm.
def process_initial_data(model_forms, initial_dict):
Nevertheless, I am able to process the form through the standard practice of using request.POST when the form is POSTed. And I don't have to pass around all kinds of information in the URL query string or modify it with JavaScript as the user enters information.

Related

Dynamically add form to formset in Django and submit with AJAX

I have read a lot of answers relating to how to dynamically add forms to an model formset in Django and can successfully implement that. However, I would now like to submit the formset with AJAX. This is mostly working now but I have an issue that I can't find a solution to in any other answer:
If you dynamically add a form to the formset, you give it a new form id number that is one larger than the maximum the form currently has and you also increment the management TOTAL_FORMS count by one. The newly added form then saves successfully as a new object.
I am trying to submit by AJAX so the user can continue editing without having the page refresh. The formset saves fine but any dynamically added forms are now existing objects. To account for this I need to increment the INITIAL_FORMS count on the management form when the save is successful. Easy enough. However, I've also realised I need to give the newly created objects an ID since they now exist in the database.
How can I get my view to tell me the ID of the new objects in its response to the AJAX call? Or is there a better way of looking at this?
Django forms and formsets are intended for classic browser-based posting of data. Though they can definitely be made to work with Javascript, the more you want to part from the normal behavior, the more complex it gets.
Depending on your requirements, you might start thinking about dropping it and switch to Javascript + REST endpoint. Of course, if you need progressive enhancements and you are required to have it work without javascript, that's not an option.
In any case, you want to have a customized view for posting from JS, so that you can get the result back and parse it easily in your AJAX handler. Probably some JSON.
There are several approaches you could take.
Have your AJAX send data to a different URL. This is pertinent if you have an API or are planning to build one at some point. So your form, when submitted normally, will do its old-style processing but your AJAX will talk to the API endpoint instead.
For instance, your form send to https://example.com/myform, but your Javascript code talks to REST api at https://example.com/api/v1/mymodel/ (sending PUT, POST and DELETE requests as appropriate).
Or if you don't have an API and building one seems overkill, you may just alter your view so it formats its output differently depending on whether the data is being submitted in the regular way or using AJAX.
You'd go about it like this:
class MyFormView(.....):
def render_to_response(self, context, **kwargs):
if self.request.is_ajax():
return self.render_to_json(context, **kwargs)
return super().render_to_response(context, **kwargs)
def render_to_json(context, **kwargs):
data = {
# see below!
}
return HttpResponse(
content=json.dumps(data).encode('ascii'),
content_type='application/json',
)
This is just an outline. You need to ensure is_ajax will detect it properly (see django doc). And you need to properly build data from context: extract the things you want to send back to your JS code and put them in the dict.
You will find it's manageable if you just do this for one, maybe two views in your project, but very quickly you'll want to have a small API instead, especially given how easy it is to build one with packages such as Django REST framework.
In your view, where you save the object, AFTER the save, the object.id will contain the new id for the object, which you can return via json or however you want in your ajax response, and then yes you will need to fill that into the formset row so that it will be submitted the next time.
One thing you have to watch out for is that django expects all existing rows to be at the top of the formset, and any new rows to be at the bottom. Otherwise, the formset save will complain about missing id's. So if you're doing any kind of sorting in your javascript, you can't do that.. unless you do quite a bit of fixing of all the field names etc in the formset. The formset code uses the numbers in the management form to determine which rows to insert and which rows to update, it does not do it on the basis of whether or not an id is present. Unfortunately...

django, return to previous page after form POST submit

In my web page I have a form that is being filled with data after some ajax requests. For example when a user chooses an item from the list, a simple ajax request is sent to the database that the item has been selected (but not confirmed, yet). Then the list on the web page reloads using a simpe ajax request (just the list, not the whole page) to fetch the new item list.
I think this is more or less a classic cart implementation.
However, when the user presses submit (classic form POST submit, not ajax POST for some reasons concerning the implementation) to confirm the whole list, I would like to return to the current page. (Current page varies) Is this possible? I am using django.
Thanks.
You can supply a next GET parameter when submitting the form, similar to django.contrib.auth's login() method:
https://docs.djangoproject.com/en/dev/topics/auth/#django.contrib.auth.views.login:
<form action="/url/to/post/form/?next={{ some_variable }}">
where the variable can simply be the current URL (taken from the request) or a generated URL. In the view processing the form, simply check for a next parameter and redirect appropriately:
from django.shortcuts import redirect
if 'next' in request.GET:
return redirect(request.GET['next'])
You may be able to use the Post/Redirect/Get Design pattern (PRG). For more general information about Post/Redirect/Get please see the following: http://en.wikipedia.org/wiki/Post/Redirect/Get There are some nice process flow diagrams there.
A generic example of a view implementing PRG might look like the following:
# urls.py
urlpatterns = patterns('',
url(r'^/$', views.my_view, name='named_url'),
)
# forms.py
class MyForm(forms.Form):
pass # the form
# views.py
def my_view(request, template_name='template.html'):
""" Example PostRedirectGet
This example uses a request context, but isn't
necessary for the PRG
"""
if request.POST:
form = MyForm(request.POST)
if form.is_valid():
try:
form.save()
# on success, the request is redirected as a GET
return HttpResponseRedirect(reverse('named_url'))
except:
pass # handling can go here
else:
form = MyForm()
return render_to_response(template_name, {
'form':form
}, context_instance=RequestContext(request))
If you need to do something more interesting with the GET, reverse can take args and kwargs. Manipulate the view params, url_pattern, and reverse call to display the results you would like to see.
One additional note is that you don't have to redirect to the same view (as this example does). It could be any named view that you would like to redirect the user to.
current page is a very vague term but i am assuming you want the page that referred you to the form page, this is normally (not always) stored in the HTTP_REFERRER header of the request itself. You could try to fetch that from the request and do a redirect.

First time jQuery $.post takes an extraordinarily long time, subsequent times normal

On a webpage we have the following system of server side form validation. For example, if the user is adding date-details for an event (and an event can contain many such date-details), we call a javascript function on click of the 'Add' button like below.
validateForm('frmName','codelibrary/classes/myclass.php','validationArrName')
where:
#frmName = form name
#codelibrary/classes/myclass.php = location of class file, that contains classes and functions for server side validation
#validationArrName = Type of validation we apply
In the php script, validationArrName is defined as a list of keys (representing form fields) and values (representing the functions we will call to validate that form field).
validationArrName = array ('fieldName1'=>validationFun1,'fieldName2'=>validationFun2);
eg:
fieldName1 = email_address
validationFun1 = validateEmail()
On the html page, we call the server side validation through ajax as follows.
$.post(className,$("form[name="+formName+"]").serialize()+"&isValidate=1&validateArrayName="+validateArrayName,function(data){ ... });
If the validation function reports an error, we display an appropriate error message back on the html page.
The problem is that when we do this for the very first time (eg: after a hard refresh of the page), submitting this date-details form for validation takes a lot of time, as compared to subsequent requests.
We observed that instead of calling the codelibrary/classes/myclass.php file once, it actually refers to this file more than 10 times before jumping to the required location (validationArrName) and running that.
For subsequent requests, it works fine and refers to that file only once.
What could be the issue here? Could there be an issue with our usage of jquery submit ?
the best thing you can do is time stuff.
in javascript:
console.time('post load'):
$.post(className,$("form[name="+formName+"]").serialize()+"&isValidate=1&validateArrayName="+validateArrayName,function(data){
console.timeEnd('post load');
console.log('data');
...
});
in php, use microtime to time different part and echo them. they will be printed in the console.
It should not be cache or include related, as ajax starts a new connection each time.
Following your comments, I edit this answer:
I'm still at loss of what happens. However I see two possibilities. The first one is that you use a "flag" to validate forms or not. When you load the page, all forms flag are unset, and first submit check them all. Subsequent submits works correctly.
Another option is that the first time you submit a form, you dont event.preventDefault() on the submit click, but it's still a loosy explanation.
I would love to see how you call the $.post(...) function (how the submit button is binded, or how $().submit() is called).

Use CodeIgniter form validation in a view

I have footer view that's included on all my pages which contains a form. I would like to be able to make use of CI's form validation library to validate the form. Is that possible?
Currently the form posts back to the current page using the PHP_SELF environment variable. I don't want to get it to post to a controller because when validation fails it loads the controller name in the address bar, which is not the desired behaviour.
Any suggestions gratefully received.
Thanks,
Gaz
One way, whilst far from ideal, would be to create a "contact" function in every controller. This could be in the form of a library/helper.
CI doesn't natively let you call one controller from another, although I believe there are extensions that enable this.
Another option would be an AJAX call instead, which would allow you to post to a generic controller, validate etc whilst remaining on the current page.
In this use case, I would definitely go for an AJAX call to a generic controller. This allows you to show errors even before submitting in the origin page.
Another way (slightly more complex), involves posting your form data to a generic controller method, passing it a hidden input containing the current URL.
The generic controller method handling your form can then redirect to the page on which the user submitted the form, passing it the validation errors or a success message using flash session variables: $this->session->set_flashdata('errors',validation_errors()) might do the trick (untested)
The good thing about this is that you can use the generic form-handling method for both the ajax case (suppressing the redirect) and the non-ajax case
AJAX would be best, just like everyone else says.
I would redirect the form to one function in one controller, you could make a controller just for the form itself. Then have a hidden value with the return URL. As far as errors go you could send them back with flashdata.
Just remember to never copy paste code, it a bad practice and guarantees bugs.
//make sure you load the proper model
if ($this->form_validation->run() == FALSE){
// invalid
$redirect = $this->input->post('url');
$this->session->set_flashdata('errors',validation_errors());
redirect($redirect);
} else {
/*
success, do what you want here
*/
redirect('send them where ever');
}

Validate data from CakePHP form with jQuery (AJAX)

I would like to validate both single field and multiple field data from a CakePHP form.
The single field validation should be done on blur from each field while the multiple field validation should be done on submitting the form.
I would like to use the $validate property declared in the Model for validating data and I would like to display the errors near each field (single field validation) and on top of the form (for multiple field validation).
My main goal is to achieve this the most "caky" way (if there is one for validating data with jQuery). I couldn't find any useful advice out there and I'm asking you for some help to get this going.
One of my concerns is how shall I pass data from the form to jQuery and then to the action that does the validation and also how shall I return and display the errors, if there are any.
Thank you in advance!
I'd suggest first making sure everything works without jQuery, then use the jQuery Form plugin to submit your forms via AJAX. If you include the RequestHandler component in your AppController, you should find that your controllers distinguish automatically between AJAX and synchronous requests.
OK, so I coded my own solution to this, but I am still waiting for a more "caky" approach.
I made two generic jQuery functions, one for single field validation and one for multiple field validation. The function should grab the data from the specified form and send it to the form's action via AJAX, to a specially created controller method which will attempt to validate data and output an AJAX response ("" for validation has passed and errors for errors in validation). Then, the result is checked in the jQuery function and the default form behaviour is triggered only if the validation has passed. Otherwise, display the errors and return false; to prevent default submission.

Resources