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Flask WTForms - Retrieve form data when submitting to a different route, i.e. a different POST route than the GET route which renders the form

I would like to set up my URLs/endpoints according to REST as closely as possible, while still utilising Flask-WTForms.
I would like my form to render at GET /posts/new, and submit to POST /post.
With Flask-WTForms I can only work out how to get it to GET/POST to the same URL.
My current code looks like this:
#post_bp.route('/posts/new', methods=['GET', 'POST'])
def show_post_form():
create_post_form = CreatePostForm()
if create_post_form.validate_on_submit():
return 'success'
return render_template('create_post_form.html', form=create_post_form)
However I would like to be able to make it look something more like this, but I just can't seem to work it out:
#post_bp.route('/posts/new', methods=['GET'])
def show_post_form():
create_post_form = CreatePostForm()
return render_template('create_post_form.html', form=create_post_form)
this route only shows the form
the form submits a POST request to /post
<form action="{{url_for('shipment.C_shipment')}}" method="POST" novalidate>
the POST /post route handles the submitted form and if there are errors for example, then it redirects back to GET /posts/new:
#post_bp.route('/post', methods=['POST'])
def create_post():
create_post_form = CreatePostForm()
if create_post_form.validate_on_submit():
return "success!"
if len(create_post_form.errors) != 0:
for error in create_shipment_form.errors:
for msg in create_shipment_form.errors[error]:
flash(msg)
return redirect(url_for('shipment.show_create_shipment_form'))
i guess creating a new CreatePostForm() object here doesn't really work..
Any suggestions?

Creating a new CreatePostForm is correct as it parses the submitted form data for you. This allows you to call validate_on_submit() on the form object.
I don't think you're generating the correct URL for the form action in your HTML snippet. The argument to url_for() should be the desired endpoint (see docs) which should be <post_bp>.create_post. This would be similar to your call
return redirect(url_for('shipment.show_create_shipment_form'))
If this does not fix the issue, please provide both frontend and backend error messages you receive when trying to send the data to /post.

I want something to be executed through django

I know this question was asked before, but none worked for me. I have this code that I want it to be executed when a button is clicked and a message is passed
import time
from sinchsms import SinchSMS
number = '+yourmobilenumber'
message = 'I love SMS!'
client = SinchSMS(your_app_key, your_app_secret)
print("Sending '%s' to %s" % (message, number))
response = client.send_message(number, message)
message_id = response['messageId']
response = client.check_status(message_id)
while response['status'] != 'Successful':
print(response['status'])
time.sleep(1)
response = client.check_status(message_id)
print(response['status'])
Basically, what I need is to add an input in a template "HTML File", this input get passed to the message variable in the code above, same with the number. I can easily do that with instances, but how can the below get executed when a button is clicked from the form in the template?
I'm kinda newbie in Django and still finding my way
Here is the tutorial that explains how to make the python file, but execute it from the shell, not a django application.
I hope I was clear describing my problem and any help would be appreciated!

All you need is a form with a message field. In a view, you want to show that form and when the user press submit, you want to execute your script.
Here is some pseudo-code:
urls.py
url('^my-page/' my_views.my_view, name='my-page'),
forms.py
SmsForm(forms.Form):
message = fields.CharField(...)
my_views.py
def my_view(request):
form = SmsForm(data=request.POST or None)
if request.method == 'POST':
if form.is_valid():
send_sms(form.cleaned_data['message']) # do this last
messages.success(request, "Success")
return HttpResponseRedirect(request.path)
else:
messages.warning(request, "Failure")
return render(request, 'my_template.html', {'form': form})
Check the Django documentation about urls, views, forms and messages and proceed step by step:
get the page to load
get the form to load
get the form submission to work and simply show "Success" or "Failure"
finally, write the send_sms function (you've almost done it)

Lets start from the dust cloud.
What you are asking is mostly about how the web pages work. You need to know how to pass parameters using HTML. There are lots of ways to do it. But with django there is a pattern.
You need a url, and a view to catch any requests. Then you need to create a template and a form inside it. With this form you could create some requests to send data to your view.
To create you need to edit urls.py inside your project add an url:
urls.py
from django.conf.urls import url
from my_app.views import my_view
urlpatterns = [
...
url(r'^my_url$', my_view, name='my_view')
...
]
For more about urls please look at URL dispatcher page at documentation.
Then create your view inside your app which is my_app in my example. Edit my_app/views.py
my_app/views.py
from django.http import HttpResponse
def my_view(request):
return HttpResponse('IT WORKS!')
This way you get a working view which could be accessed with path /my_url. If you run ./manage.py runserver you could access your view from http://localhost:8000/my_url.
To create a form you need to create a template. By default django searches app directories for templates. Create a templates directory in your app, in our case my_app/templates and create an HTML file inside. For example my_app/templates/my_form.html. But i advice to create one more directory inside templates directory. my_app/templates/my_app/my_form.html. This will prevent template conflicts. You can check Templates page at documentation for more.
my_app/templates/my_app/my_form.html
<html>
<body>
<form action="/my_url" method="POST">
{% csrf_token %}
<input type="text" name="number">
<input type="text" name="message">
<input type="submit" value="Run My Code">
</form>
</body>
</html>
This is the one of the ways of creating your form. But I do not recommend it. I will make it prettier. But first lets "Make it work", edit your views.py:
csrf_token is a django builtin template tag, to put CSRF token into your form. By default django requires CSRF tokens at every post
request.
my_app/views.py
from django.http import HttpResponse
from django.shortcuts import render
def my_view(request):
if request.method == 'GET':
return render('my_app/my_form.html')
elif request.method == 'POST':
# get post parameters or None as default value
number = request.POST.get('number', None)
message = request.POST.get('message', None)
# check if parameters are None or not
if number is None or message is None:
return HttpResponse('Number and Message should be passed')
# your code goes here
...
return HttpResponse('Your code result')
Till this point the purpose of this answer was "Making it work". Lets convert it nice and clean. First of all we would create Form. Forms are like models, which helps you create forms as objects. It also handles form validations. Forms are saved inside forms directory generally. Create my_app/forms.py and edit it:
my_app/forms.py
from django import forms
class MyForm(forms.Form):
number = forms.CharField(max_length=15, required=True)
message = forms.CharField(max_length=160, required=True)
Put your form inside your template:
my_app/templates/my_app/my_form.html
<html>
<body>
<form action="{% url 'my_view' %}" method="POST">
{% csrf_token %}
{{ form }}
</form>
</body>
</html>
Besides the form, the action of the HTML form tag is also changed.
url template tag is used to get url form url name specified in urls.py.
Instead of url tag, {{ request.path }} could have been used.
Create a form instance and pass it to the template rendering:
my_app/views.py
from django.http import HttpResponse
from django.shortcuts import render
from .forms import MyForm
def my_view(request):
if request.method == 'GET':
form = MyForm()
return render('my_app/my_form.html', {'form': form})
elif request.method == 'POST':
form = MyForm(request.POST)
# check if for is not valid
if not form.is_valid():
# return same template with the form
# form will show errors on it.
return render('my_app/my_form.html', {'form': form})
# your code goes here
...
return HttpResponse('Your code result')
You can use class based vies to write your view, but it's not necessary. I hope it helps.

You can create a view that takes up query parameters from the url and use it for further implementation. Then you can create a link/button in the html template which can redirect you to that url. For example:
in urls.py:
url(r'^run_a/(?P<msg>\w{0,25})/(?P<num>\w{0,25})/$', yourcode, name='get_msg'),
in template:
submit
in views.py:
def get_msg(request,msg,num):
message=msg
number=num
#rest of the code
Hope this helps :)

Flask putting form into URL

I've been working on a form that sends data to a scraper and simultaneously generates a URL from form input. The returned templates works flawlessly, but the URL change ends up giving me the entire form in the URL and I can't figure out why.
The URL ends up looking like this:
http://localhost/options/%3Cinput%20id%3D%22symbol%22%20name%3D%22symbol%22%20type%3D%22text%22%20value%3D%22%22%3E
I'd like it to look like this:
http://localhost/options/ABC
Form class:
class OptionsForm(Form):
symbol = StringField('Enter a ticker symbol:', validators=[Required(), Length(min=1, max=5)])
submit = SubmitField('Get Options Quotes')
Views:
# Where the form data ends up
#app.route('/options/<symbol>', methods=['GET', 'POST'])
def options(symbol):
# Created this try/except so I could test functionality - for example, I can do 'localhost/options/ABC' and it works
try:
symbol = request.form['symbol']
except:
pass
return render_template('options.html', symbol=symbol, company_data=OS.pull_data(symbol, name=True))
# Where the form lives
#app.route('/', methods=['GET', 'POST'])
def index():
form = OptionsForm()
print(form.errors)
if form.validate_on_submit():
return redirect(url_for('options', symbol=form.symbol.data))
return render_template('index.html', options_form=form)
Template:
<div id="options_div">
<form method="POST" name="symbol_form" action="{{ url_for('options', symbol=options_form.symbol) }}">
{{ options_form.hidden_tag() }}
{{ options_form.symbol(size=10) }}
{{ options_form.submit(size=10) }}
</form>
Any help would be appreciated.

Try adding enctype='multipart/form-data' to the form tag. It looks like your form is using application/x-www-form-urlencoded, the default.
Edit OK so check this out. When your template is being rendered there is no value in that data attribute (In the url_for call). When not referencing the data attribute (as your original question shows), you're referencing the actual form element (which is why you see all of that html being passed in the url). Here are your options (that I see):
Use some kind of frontend javascript to bind the form's action attribute to the value in the input box. Something like angular would help for this (but is overkill if you don't use any of its other features).
Just have the form POST to /options (no symbol in url). Then, grab the symbol attribute from the form data.

Keeping typed in form data in django

I have a form where i would like the data to be still present in the form fields if there is a form validation error. By default when i try to submit data in the form django clears all the fields. What is the best approach for doing this using django 1.6 ?
I thought i would just fetch data like so: request.POST['field'], send it to the template and then let the template run a conditional if statement by im getting an error from django telling me that the values isnt found in the multidict but it is found when i fill out the form field so that doesnt seem to work. I also tried to check if the value was present directly in the view but i ended up with a ridicoulus amount of if statements to be able to do this which is just an ugly hack.
Can someone suggest a good working solution on this ? Ive seem threads on SO on kind of the same problem but people just wrote that it should be the default to keep data but that doesnt seem to be the case for django 1.6.

I imagined your form class MyForm(). If the form is not valid send form = MyForm(request.POST) to template.
some view:
def myview(request):
if request.method=='POST':
form = MyForm(request.POST)
if form.is_valid():
form.save()
return render(request, 'success.html')
else:
form = MyForm()
return render(request, 'form.html',{'form':form})
<form action="." method="POST">
{{form}}
</form>

Do we need CSRF verification for every POST request?

I am just building a simple HTML form with POST method and unfortunately I am finding CSRF verification error.
This is just a simple html form using POST method on localhost. There are no cross sites involved. I could definitely fix it by using csrf_token but I still don't understand why django is asking me for that..
There are no re-directions/ iframes involved here...
So, why this is happening?? is this normal to all ??
# Also tried using RequestContext(request) but there isn't any change in the error
#settings.py
'django.middleware.csrf.CsrfViewMiddleware' in MIDDLEWARE_CLASSES
#views.py
# url for home page is "" i.e, http://127.0.0.1:8000/
def HomePage (request):
if request.method == "POST":
form = myForm(request.POST)
if form.is_valid():
data = form.cleaned_data
context = { "myForm" : myForm(choices),
"values" : data,
}
return render_to_response("home.html", context)
else:
form = myForm(choices)
context = {"myForm" : form}
return render_to_response("home.html", context)
# home.html
<div id="pingmeeForm">
<form action="" method="post">
<table>
{{myForm.as_table}}
</table>
<input name="enter" type="submit" value="enter"/>
</form>
{{values}}
</div>
# forms.py
class myForm (forms.Form):
def __init__(self, my_choices,*args, **kwargs):
super(myForm, self).__init__(*args, **kwargs)
self.fields['Friends'] = forms.ChoiceField(choices=my_choices)
message = forms.CharField()

If you do a post request, you typically change the state of the server. If you change the state of the server, you don't want to allow other sites to do so. To protect against other sites issueing post-requests on your server, you add csrf protection. Therefore the solution should (imho) never be to remove the Csrf protection. Depending on the situation, either of the following two is the case:
Your post request does not change the state. In that case, make it a get request.
Your post request changes the state. You need CSRF.

The error message you got but didn't show explains exactly what you are doing wrong: you should ensure that
the view function uses RequestContext
for the template, instead of Context.