CODEWITHSUNDEEP
pythondjango
user7441730
user7441730

Reputation:

Django html onclick to open another html page

i am currently facing problem on opening html page using django although i have tried to figure out on the url.py, view.py, and the html page. My code is stated below:

batterycurrent.py under views folder

from __future__ import absolute_import
from __future__ import unicode_literals

from django.core.files.storage import default_storage
from django.core.urlresolvers import reverse_lazy
from django.http import HttpResponseRedirect
from django.utils import timezone
from django.views.generic import FormView, TemplateView
from sendfile import sendfile
import os.path

from .mixin import AjaxTemplateMixin, PermissionRequiredMixin, PageTitleMixin
from ..forms import DiagnosticsForm
from ..tasks import dump_diagnostics
from django.shortcuts import render

class DiagnosticMixin(PageTitleMixin, PermissionRequiredMixin):
    permission_required = ['system.view_log_files']
    page_title = 'Diagnostics'
    form_class = DiagnosticsForm

class BatteryCurrentView(DiagnosticMixin,FormView):
     template_name = 'system/batterycurrent.html' 

    def batterycurrent(request):
    return render(request, 'system/batterycurrent.html')

url.py

from __future__ import absolute_import
from __future__ import unicode_literals

from collections import OrderedDict
from django.conf.urls import url
from django.core.urlresolvers import reverse_lazy
from .views import BatteryCurrentView

sub_urlpatterns['diagnostic'] = [
    url(r'^diagnostics$', DiagnosticView.as_view(), name='diagnostic'),
    url(r'^diagnostics/download', DiagnosticDownloadView.as_view(), name='diagnostic-download'),
    url(r'^diagnostics/batterycurrent', BatteryCurrentView.as_view(), name='batterycurrent'),
]

**the batterycurrent.html is within diagnostic.html.

batterycurrent.html(Template (the html is in the system folder)

<li><a href="{% url 'system:batterycurrent' %}">Battery Current Vs Timestamp</a></li>

when i started to execute the code, the errors appeared 

i) importError batterycurrentView couldn't be imported
ii)Reverse for 'batterycurrent' with arguments '()' and keyword arguments '{}' 
   not found. 0 pattern(s) tried: []

Please guide me on this

Upvotes: 0

Views: 1433

Answers (3)

zaidfazil
zaidfazil

Reputation: 9235

Try it again after adding some context in the view, or leave it empty.

def batterycurrent(request):
    return render (request, template, {})

Upvotes: 0

Ahmed Ginani
Ahmed Ginani

Reputation: 6650

Seems like the problem is in import the module in url.py file.

If you have created BatteryCurrentView module inside the batterycurrent.py and your url.py file in the different folder then you need to import your module through the name of the folder with the file name.

So Your file is inside the view folder then you should use

from views.batterycurrent import BatteryCurrentView

And if in case your file url.py resides in the same folder inside view then you need to just import module like this:

from .batterycurrent import BatteryCurrentView

Upvotes: 0

Daniel Roseman
Daniel Roseman

Reputation: 599490

You haven't provided your full app layout so this is a bit hard to answer, but it sounds like you have a views folder containing a file "batterycurrent.py", which in turn contains your view BatteryCurrentView.

If so, as well as ensuring that you have an (empty) __init__.py file in that directory, you urls.py needs to import the view from the module it is in:

from views.batterycurrent import BatteryCurrentView

Also, I don't understand what you are doing with sub_urlpatterns in that file. You don't define that dictionary anywhere; but in any case, the patterns must be in a list called just urlpatterns.

Finally, ensure that that file is actually called urls.py, not url.py.

Upvotes: 1

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