I am a newbie in Django, though I code in python using Twisted, and I still have a long way to go dealing with this platform. Ihave a problem regarding the use of url templates in forms. I have defined a regex path
(url(r'^search/(?P<key>\w*)/$', views.searchKey, name='search'),)
in my urls.py. The url path works when I hard code the url path. For example:
"/search/potatoes/"
But when I use forms to post a search to that path, I always get "/search//". Below is my forms code. What seems to be happening?
<form action="{% url 'feeds:search' key %}" method="post">
{% csrf_token %}
<input type="text" name="key" id="key" />
<input type="submit" value="Filter" />
</form>
Thanks in advance!
You've set that as a keyword argument, so I think you'll need to use {% url 'feeds:search' key='potatoes' %} or if you have an actual variable there {% url 'feeds:search' key=key_variable %}.
Now if you're trying to get the form to submit to that URL you'll need to do some JavaScript on the frontend to get that URL changed based on what's input into the key field.
Related
I have a list of tuples in this format: [(a,b,c), (d,e,f),(g,h,i)]. For some reasons, I need to pass it as a hidden field in my form and in my views I need to retrieve it. However, Im not able to do it.
Here is my code snippet:
html template
<form id="form1" action="{% url 'job_recommendations' %}" method="POST">
{% csrf_token %}
<a onclick="document.getElementById('form1').submit();">View more recommendations </a
<input type="hidden" name="recommendations" value="{{ job_recommendations }}"> #it
is the list that I want to pass. It already has correct values
</form>
views.py
def job_recommendations(request):
if request.method == "POST":
recommendations = request.POST['recommendations']
for job, recruiter, percentage in recommendations:
print(percentage)
return render(request, 'recommendations.html')
It's not the best way to do this, but you can use ast module.
Jinja would put [(a,b,c), (d,e,f),(g,h,i)] as string in template. You would get this string in your view, then you can call ast.literal_eval(str) function on this string to convert this string into list.
Please remember that this is very unsecure because it allows user from frontend to execute code on backend.
I was searching for this answer but none met my expectation. So, In my template I have some content and wanted to add button (which later will add to favorites). After clicking I want to call method from my views.py and redirect to other view.
my views.py
def home(request):
//logic here
request.session['url'] = url
return render(request,'file.html')
def function_to_call(request):
///logic here
url = request.session.get('url')
return render(request,'second_file.html',url=url)
file.html
<form action="{% url 'function_to_call' %}">
<button id="submit" type="button" value="Click" />
</form>
and in my urls.py
url(r'^function_to_call/',views.function_to_call,name='function_to_call'),
Unfortunately, after clicking on button, nothing happens
unless you are submitting a form, you should use
Click
If for some reason you need to use a POST request rather than a GET this will work:
<form method="POST" action="{% url 'function_to_call' %}">
<button id="submit" type="submit" value="Click" />
</form>
Using a post can be helpful when you don't want to include data in the querystring because it's a little less secure than having the parameters in the request's body.
I am use Django 1.8 and Python 3.5.2, and try recive POST data from simple html form
<form action="/shop/order" method="POST">
{% csrf_token %}
<input type="hidden" name="product-code" value="{{ product.product_code }}">
<input type="text" name="email">
<input type="submit">
</form>
It view where I try get need for me data, but QueryDict empty for POST, when I change POST to GET in form and view all work.
def order_product(request):
test = request.POST.get('product-code', '')
mail = request.POST.get('email', '')
# logger.info(test)
return render(request, 'shop/test.html', dict(test=test, email=mail))
And I'am not understand why is that. Similarly I try parse(decode binary to utf-8 etc) body and I recive empty string.
urls.py
url(r'^order/?$', views.order_product),
Thanks all, the moral of this fable is as follows, use name argument in urls, and use {% url 'name' %} in template. I have i18n in urls(from djangoCMS) and when I harcode url in form isn't work for POST method.
Hi I got a simple form for a POST request and it works when I'm only having one input, but not two inputs together. Can someone show me some light on this?
index.html
<form name="input" action="{% url 'sending' %}" method="post">
{% csrf_token %}
Recipient: <input type="text" name="recipient">
<br>
Message: <input type="text" name="content">
<br>
<input type="submit">
</form>
views.py
def sending(request):
recipient = request.POST.get('recipient','')
content = request.POST.get('content','') #not working when I am doing this...
someMethod(recipient, content)
return HttpResponseRedirect(reverse('results'))
Adding a "forms" portion to your setup will help you greatly... see the quickstart docs on forms here: https://docs.djangoproject.com/en/1.6/topics/forms/
In particular, check out "using a form in a view": https://docs.djangoproject.com/en/1.6/topics/forms/#using-a-form-in-a-view
Basically, you end up with a "forms.py" file which defines your form fields. Then, after it all processes, you get a simplier API into your form fields that looks like this:
form.cleaned_data['recipient']
form.cleaned_data['content']
etc.
I have this in my views.py file as the view config for my home page:
#view_config(route_name='home_page', renderer='templates/edit.pt')
def home_page(request):
if 'form.submitted' in request.params:
name= request.params['name']
body = request.params['body']
page=Page(name,body)
DBSession.add(page)
return HTTPFound(location=request.route_url('view_page',pagename=name))
return {}
Also, here is the form in the edit.pt template:
<form action="/view_page" method="post">
<div>
<input type="text" name="name"/>
</div>
<div>
<input type="text" name="body"/>
</div>
<label for="stl">Stl</label>
<input name="stl" type="file" value="" />
<input type="submit" name='form.submitted' value="Save"/>
</form>
Also in my init.py file I have
config.add_route('home_page', '/')
config.add_route('view_page', '/{pagename}')
right now when I submit the form it just tries to go to localhost:6543/view_page. This returns a 404 as there is no view_page resource or route leading to it. Instead I want it to go to localhost:6543/(the name of the page I just created aka the first input box in the form). How can I do this?
Edit: I am worried that something else may be telling it to route to view_page because I even tried changing it to
return HTTPFound(location=request.route_url('front_page',pagename=name))
And it still goes to /view_page. There is no route named front_page, so I would at least suspect it to throw an error.
Also, I would really appreciate it if you could tell me where you found the info. I have been looking at http://docs.pylonsproject.org/projects/pyramid/en/1.4-branch/api/request.html?highlight=request.route_url#pyramid.request.Request.route_url but can't seem to find use from it.
Edit: should I be using an asset specification instead of a path name? so
return HTTPFound(Location=request.route_url('tutorial:templates/view.pt','/{pagename}'))
Also, I am working through this article which seems very helpful with the syntax: http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/urldispatch.html#urldispatch-chapter
I think your form should submit to "/", ie.
<!-- where your home_page route is waiting for the POST -->
<form action="/" method="post">
With the prior answers this now looks correct:
return HTTPFound(location=request.route_url('view_page', pagename=name))
My first guess is that it's location not Location as the argument to HTTPFound.
from the link you give
it's should be
return HTTPFound(location=request.route_url('view_page',pagename=name))
when you had add this route
config.add_route('view_page', '/{pagename}')
and set the variable name before
name= request.params['name']