Use different SQL data depending on url in Django

Question

If i have a url that is "someurl/report/1/" where "report" is an app and "1" corresponds to a certain id of a model in my SQL database, how would i substitute the variables in the template with that specific models data?

In my case I am writing a website that displays a surf report for different beaches. I have set it up so each SQL model id corresponds to a different beach. So if I wanted to use the data of beach "3" in the template, how would i display those in the html template?

TRACEBACK

Using the URLconf defined in surfsite.urls, Django tried these URL patterns, in this order: ^admin/ ^report/ ^$ [name='index'] ^report/ (?P[0-9]+)$ [name='get_report'] The current URL, report/1/, didn't match any of these.

#URLS.PY
from django.conf.urls import url
from django.conf.urls.static import static
from . import views

    urlpatterns = [
        # /index/
        url(r'^$', views.index, name='index'),

        # /report/
        url(r'(?P<beach_id>[0-9]+)$', views.get_report, name='get_report'),
    ]


#MY TEMPLATE
<!DOCTYPE html>
<html>
    {% load staticfiles %}
    <link rel="stylesheet" type="text/css" 
    href="{% static 'report/css/style.css' %}"/>

    <link href="https://fonts.googleapis.com/css?family=Biryani:300,400,800" 
    rel="stylesheet"/>

   <head>
      <title>REALSURF</title>
  </head>

   <body>

      <h1>
        <form id="search">
          <input id="search" type="text">
        </form>
      </h1>

    <div class="container">
      <div class="column">
        <div class="text">Wind</div>
        <div class="number">{{Break.wind}}</div>
      </div>            
      <div class="column">
        <div class="text">Wave Height</div>
        <div class="number">{{Break.low}}-{{Break.high}}</div>
      </div>    
      <div class="column">
        <div class="text">Tide</div>
        <div class="number">{{Break.tide}}</div>
      </div>    
    </div>

    <h2>
      REALSURF
    </h2>

    <h3>
      A simple site by David Owens
    </h3>

   </body>
</html>




#MY MODEL
from __future__ import unicode_literals   
from django.db import models

class Beach(models.Model):
    name = models.CharField(max_length=50)
    high = models.SmallIntegerField()
    low = models.SmallIntegerField()
    wind = models.SmallIntegerField()
    tide = models.DecimalField(max_digits=3, decimal_places=2)

    def __str__(self):
        return self.name


#MY VIEWS
from django.http import Http404
from django.shortcuts import render
from .models import Beach

def index(request):
    allBeaches = Beach.objects.all()
    context = {
        'allBeaches': allBeaches,
    }
    return render(request, 'report/index.html', context) 

def get_report(request, id):
    try:
        beach = Beach.objects.get(id=id)

    except Beach.DoesNotExist:
        raise Htpp404("404")

    return render(request, 'report/index.html', {'beach': beach})

Zagorodniy Olexiy · Accepted Answer · 2016-12-22 23:51:24Z

1

If i understand you right, you want to create view, that render beach separately You can do something like this, your view:

def get_beach(request, id)
  beach = Beach.objects.get(id=id)
  return render(request, 'path/to/your/template', {'beach':beach})

urls:

url(r'^someurl/report/(?P<id>[0-9]+)$', views.get_beach(), name='get_beach'),

template url to this page:

<a href= "someurl/report/{{beach.id}}">beach</a>

Edited Its your view, as i understand, this is detail view, but select all objects(Break.objects.all())

def detail(request, break_id):
    try:
        allBreaks = Break.objects.all()

    except Break.DoesNotExist:
        raise Http404("404")

    return render(request, 'report/index.html', {'allBreaks': allBreaks})

so you have to change this on this:

 def detail(request, break_id):
      try:
        break_detail = Break.objects.get(id=break_id)
        return render(request, 'path/to/your/template', {'break_detail':break_detail})
      except Break.DoesNotExist:
        raise Http404("404")

then your url should look like this:

url(r'^someurl/report/(?P<break_id>[0-9]+)$', views.detail(), name='detail'),

or you can use get_object_or_404:

def detail(request, break_id): break_detail = get_object_or_404(Break, id=break_id) return render(request, 'path/to/your/template', {'break_detail':break_detail})

Url is the same.

So if you want to access to wind field, you just write this tempalte tag {{break_detail.wind}}

UPD2

change place from this

^admin/
^report/ ^$ [name='index']
^report/ (?P<beach_id>[0-9]+)$ [name='get_report']

to this:

^admin/
^report/(?P<beach_id>[0-9]+)$ [name='get_report']
^report/^$ [name='index']

and delete spaces in urls after report/

edited Dec 22, 2016 at 23:51

answered Dec 22, 2016 at 22:55

Zagorodniy Olexiy

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8 Comments

David Owens Over a year ago

the line beach = Beach.objects.get(id=id) is what filters to get the exact beach from the URL correct?

David Owens Over a year ago

So how would i access the wind variable, for example, in the template?

Zagorodniy Olexiy Over a year ago

its hard to answer on your question, because you didn't post you models and views. But if model Beach has field wind, you can access to it {{beach.wind}}. If you want more specific answer you have to post more information.

David Owens Over a year ago

The model and views are in the bottom of the code section if you scroll down. I am getting a syntax error on the url line you gave when i try to run the server?

David Owens Over a year ago

now when testing the site, "someurl/report/" works but any url like "someurl/report/1/" fails.

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