passing parameters through url to view django

Question

I'm having some difficulty passing values to my view through url.

So far, I've managed to run my view without any issues.

view.py (draws a graph):

def draw(request)
    ....
    ....
    return HttpResponse (buffer.getvalue(), content_type="Image/png")

but I need my view to take input from users, so I edited it and added an extra parameter:

def draw(request, loan_amount)
     loanAmount = loan_amount
     .....

The user input is passed from a form, to another view:

def search_member(request):
     loanAmount = request.GET.get('desired_loan')
     return render(request, 'blog/search_member.html', {'loanAmount':loanAmount)

In my template, I insert the user input in the draw's parameter:

<img src="http://test.com/graph/{{ loanAmount }}">

This is suppose to draw an image base on the user input, instead I get no image at all.

If I remove the parameter, the image works fine. I'm assuming I'm doing something wrong with the setup of the parameter, most likely in the template, or url:

url(r'^graph/(?P<desired_loan>\d+)/$', views.draw, name='draw'),

I have tested the view, form and everything else, they all work. How can I narrow down this problem to find the solution?

Any direction/help would be appreciated,

thanks,

<form method="GET" action="/search_member/" class="navbar-form pull-left">

            <input type="number" step = "any" id="searchBox" class="input-medium search-query" name="desired_loan" placeholder="Desired Loan"><br>
            <input type="number" step = "any" id="searchBox" class="input-medium search-query" name="repayment_time" placeholder="Payment Time"><br>

            <input type="submit" class="btn" value="Draw Graph" >

        </form><br>
        <img src="http://test.com/graph/{{ loanAmount }}">

screen shot:

@NabeelAhmed Hi Nabeel, why am I able to load <img src="test.com/graph"> but not <img src="test.com/graph{{ loan_amount }}"> ?? Meanwhile I will read about loading static. — almost a beginner
– almost a beginner, Commented Aug 22, 2016 at 12:11

Anna · Accepted Answer · 2016-08-22 12:40:54Z

3

Fo url like

url(r'^graph/(?P<desired_loan>\d+)/$', views.draw, name='draw'),

Change your view to:

def draw(request, *args, **kwargs):
 loanAmount = kwargs['desired_loan']
 .....

Then change in template:

<img src="{% url 'graph' loanAmount %}">

edited Aug 22, 2016 at 12:40

answered Aug 22, 2016 at 11:52

Anna

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4 Comments

Anna Over a year ago

If you use {% url 'graph' loanAmount %} what error will you have? And change def draw(request, loan_amount) to def draw(request, *args, **kwargs)

almost a beginner Over a year ago

In your latest update, I get 2016-08-22 12:29:43,515 :Not Found: /static/path/to/user image/10000/img.png. I have made the necessary changes. Can you please change your code to before your update.

almost a beginner Over a year ago

I get the following error, just read the error log: ymin = (-desiredLoan - (repaymentTime * 25)) TypeError: bad operand type for unary -: 'str'

Anna Over a year ago

loanAmount = int(kwargs['desired_loan'])

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