Django URL/Views extra parameters

Question

In Django1.6, is there a way to pass a dynamic parameter to my view or URL without parsing the URL?

Ideally I would want a urls.py that looks like:

url(r'^dash/$',
    dash_view.account_modify,
    {'account': **dynamic_account_identifier_here**}
    name='dash_account_modiy')

And in views.py:

def account_modify(request, account, 
                   template_name='profile.html, 
                   change_form=AccountModifyForm):
    ...

:param account:
comes from model:

class Dash(models.Model):
    name = models.Charfield()
    account = models.IntegerField()
    ....

Basically, I would really like to avoid a urls.py with the account identifier as part of the string, like:

url(r'^dash/(?P<account>\w+)/$',
    dash_view.account_modify,
    name='dash_account_modiy')

Any suggestions on how I can pass these values from the a template to the processing view for use in the AccountModifyForm (which expects that 'account' parameter)?

Alasdair · Accepted Answer · 2013-11-22 23:20:07Z

3

url(r'^dash/$',
    dash_view.account_modify,
    {'account': **dynamic_account_identifier_here**}
    name='dash_account_modify')

You can't dynamically evaluate anything there, because the dictionary is evaluated only once, when the URL conf is loaded.

If you want to pass information from one view to another your three options are:

in the URL, which you don't seem to want to do
as GET or POST data
store it in the session in one view and retrieve it from the session in the next

answered Nov 22, 2013 at 23:20

Alasdair

310k59 gold badges605 silver badges534 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

parsenz Over a year ago

Yeah...just figured that out...thanks for the pointer still. Had a bit of a moment.

Alasdair Over a year ago

Glad you figured it out :)

parsenz · Accepted Answer · 2013-12-18 02:44:01Z

1

If anyone cares...figured it out...

In the template:

{% for dash in dashes %}
     blah blah blah
     <form action="..." method="POST">
         <input type="hidden" name="id" value="{{ dash.account }}">
         {{ form.as_ul }}
         <input type="submit" value="Do stuff">
     </form>
{% endfor %}

In the views:

if request.method == 'POST'
    account = request.POST['id']
    # be sure to include checks for the validity of the POST information
    # e.g. confirm that the account does indeed belong to whats-his-face
    form = AccountModifyForm(request.POST, account,
                             user=request.user)
    ....

edited Dec 18, 2013 at 2:44

answered Nov 22, 2013 at 23:22

parsenz

4095 silver badges10 bronze badges

1 Comment

Alasdair Over a year ago

Be careful, the user can modify the value of the hidden field and edit any account, which might be a security issue in some cases.

Collectives™ on Stack Overflow

Django URL/Views extra parameters

2 Answers 2

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related