Django - How to add a link to a table that will open another page(django app) from an html template and pass a variable with it.

[Patrick Carra]

Hello I have an app that displays some database information in a table.  Inside of the html template I am making an edit link that I want to open another app(page viewLit) while passing a value to it's view.  I have added my code below.  My question is I am unsure of how to make this links url and pass the object data located inside circuit.circuitid along with it.  I haven't been able to find the right way to code this yet and this is just how I thought that this should be done. If anyone has a better idea I am open to suggestions.

search_custom.html(code for link)

{% for circuit in filter.qs %}

<tr>

    <td class="actions">

        <a href="" class ="view-item" title ="View">View</a>

    </td>

    <td>{{ circuit.circuitid }}</td>

</tr>

{% endfor %}

myapp/myapp/urls.py

urlpatterns = [

    path('viewLit/', include('viewLit.urls')),

]

myapp/viewLit/urls.py

urlpatterns=[

    path('viewLit/circuitid.id', views.viewLit, name='viewLit'),

]

views.py

def viewLit(request, circuitid):

    #display records fields here

    return HttpResponse("You are at the viewLit page!")

[Randy Hughes]

views.py

class viewLit(TemplateView):

def get(self, request, strategy=None):

circuit_id = self.request.query_params.get('circuitid', None)

urls.py

path('viewlit/', viewlit.as_view(), name='viewlit')

search_custom.html

<td>

<a href="{% url viewlit %}?circuitid={{ circuit.circuitid }}">

{{ circuit.circuitid }}

</a>

</td>

''' this is an example using a query parameter

the url in urls.py does not need to include this parameter as a slug

the view is able to get the parameter by key name (the same way you get a key from a dict)

the <a></a>tag is formatted to create the url ex (127.0.0.1/foo/) query parameters are everything following the ?

so as you can do something like this 127.0.0.1/foo/?field1=a&field2=b&field3=c

I am not sure if this is the better than your solution but this should work. As well you can add a redirect in viewlit to a different page if

circuit_id does not exist.

'''

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[Patrick Carra]

Thank you Randy!

[Patrick Carra]

This worked for me but upon more testing I discovered some values such as WAVE/123456//IMP get changed to WAVE/123456//IMP during the process of passing this parameter as a url and django interpreting it.  As a result I get an error: Circuitinfotable matching query does not exist.  I'm assuming this is because django is interpreting the second / as an escape character?  How do I get it to include all /'s in the query so that a correct result will be returned.

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