Reverse for 'edit' with arguments '(9,)' and keyword arguments '{}' not found. 0 pattern(s) tried: []

By : wemode
Source: Stackoverflow.com
Question!

I'm getting an error when trying to access edit link in django, i have looked here on stack overflow but i haven't found the solution that works in my case.

ERROR : Exception Type : NoReverseMatch Exception Value : Reverse for 'edit' with arguments '(9,)' and keyword arguments '{}' not found. 0 pattern(s) tried: [] arguments '{}' not found. 0 pattern(s) tried: []

this is my urls.py

from django.conf.urls import url, include
from django.contrib import admin
from posts import views

urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^create/$', views.create, name='create'),
    url(r'^(?P<id>\d+)/$', views.show_post, name = 'show_post'),
    url(r'^(?P<id>\d+)/edit/$', views.update_post, name = 'update_post'),
    url(r'^(?P<id>\d+)/delete/$', views.delete_post),
]

views.py

from django.contrib import messages
from django.shortcuts import render, get_object_or_404, redirect
from django.http import HttpResponse, HttpResponseRedirect

from .forms import PostForm
from .models import Post

# Create your views here.

def index(request):
    post_list = Post.objects.order_by('-created_date')[:10]
    context = {'post_list': post_list}
    return render(request, 'index.html', context)

def create(request):
    form = PostForm(request.POST or None)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
        #flass messages
        messages.success(request, "Successfully created")
        return HttpResponseRedirect(instance.get_absolute_url())


    context = {
        "form":form,
    }
    return render(request, 'post_form.html', context)   


def show_post(request, id=None):
    instance = get_object_or_404(Post, id=id)
    context = {'instance': instance}
    return render(request, 'show_post.html', context)


def update_post(request, id=None):
    instance = get_object_or_404(Post, id=id)
    form = PostForm(request.POST or None, instance=instance)
    if form.is_valid():
        instance = form.save(commit=False)
        instance.save()
        messages.success(request, "Post updated")
        return HttpResponseRedirect(instance.get_absolute_url())
    context = {
        "form":form,
        "instance":instance
    }
    return render(request, 'post_form.html', context)


def delete_post(request, id=None):
    instance = get_object_or_404(Post, id=id)
    instance.delete()
    messages.success(request, "Successfully deleted")
    return redirect("posts:index")

show_post.html

{% extends "base.html" %}

<div class="container">
{% block content  %}
    <h1> {{instance.title}} </h1>
    <h3>{{instance.content| linebreaks}} </h3>
    <a href="{% url 'posts:index' %}"> Home</a> | <a href="{{instance.url}}" target="_blank" > visit url</a> | 
    <a href="{% url 'posts:update_post' %}"> Edit</a> 

{% endblock %}

</div>
By : wemode


Answers

You need to use the namespace. Rather than 'edit', you should use 'posts:edit'.

Or 'posts:update_post' depending which name you're using in urls.py.



There is a standard trick for computing the Fibonacci numbers that can easily be adapted to your problem. The naive definition for Fibonacci numbers is:

fibFunction :: Int -> Integer
fibFunction 0 = 1
fibFunction 1 = 1
fibFunction n = fibFunction (n-2) + fibFunction (n-1)

However, this is very costly: since all the leaves of the recursion are 1, if fib x = y, then we must perform y recursive calls! Since the Fibonacci numbers grow exponentially, this is a bad state of affairs to be in. But with dynamic programming, we can share the computations needed in the two recursive calls. The pleasing one-liner for this looks like this:

fibList :: [Integer]
fibList = 1 : 1 : zipWith (+) fibList (tail fibList)

This may look a bit puzzling at first; here the fibList argument to zipWith serves as the recursion on two indices ago, while the tail fibList argument serves as the recursion on one index ago, which gives us both the fib (n-2) and fib (n-1) values. The two 1s at the beginning are of course the base cases. There are other good questions here on SO that explain this technique in further detail, and you should study this code and those answers until you feel you understand how it works and why it is very fast.

If necessary, one can recover the Int -> Integer type signature from this using (!!).

Let's try to apply this technique to your function. As with computing Fibonacci numbers, you need the previous and second-to-last values; and additionally need the current index. That can be done by including [2..] in the call to zipWith. Here's how it would look:

waves :: [Integer]
waves = 1 : 1 : zipWith3 thisWave [2..] waves (tail waves) where
    thisWave n back2 back1 = ((3 * n - 3) * back2 + (2 * n + 1) * back1) `div` (n + 2)

As before, one can recover the function version with (!!) or genericIndex (if one really needs Integer indices). We can confirm that it computes the same function (but faster, and using less memory) in ghci:

> :set +s
> map wave [0..30]
[1,1,2,4,9,21,51,127,323,835,2188,5798,15511,41835,113634,310572,853467,2356779,6536382,18199284,50852019,142547559,400763223,1129760415,3192727797,9043402501,25669818476,73007772802,208023278209,593742784829,1697385471211]
(6.00 secs, 3,334,097,776 bytes)
> take 31 waves
[1,1,2,4,9,21,51,127,323,835,2188,5798,15511,41835,113634,310572,853467,2356779,6536382,18199284,50852019,142547559,400763223,1129760415,3192727797,9043402501,25669818476,73007772802,208023278209,593742784829,1697385471211]
(0.00 secs, 300,696 bytes)


With n=30, you need to compute wave 29 and wave 28, which, in turn, needs to compute wave 28, wave 27 twice and wave 26 and so forth, this quickly goes in the billions.

You can employ the same trick that is used in computation of the fibonacci numbers:

wave 0 = 1
wave 1 = 1
wave n = helper 1 1 2
    where
       helper x y k | k <n      = helper y z (k+1)
                    | otherwise = z
                    where z = ((3*k-3) * x + (2*k+1) * y) `div` (k+2)

This runs in linear time, and the helper has, for every k the values for wave (k-2) and wave (k-1) ready.

By : Ingo


This video can help you solving your question :)
By: admin