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Python Type Error when setting up pagination in Django

Hi I am following the official documentation and setting up pagination for my website, my index template looks like this;

{% for post in list_of_posts %}
<div class="body"><a class="title" href="/post/{{post.id}}"><h2>{{ post.title }}</h2></a>
<P>{{ post.body|truncatewords:50|wordwrap:110 }}</P>    
<h3>{{ post.date|date:"jS F Y" }}</h3>
<hr>
</div>
{% endfor %}
<div class="pagination">
<span class="step-links">
    {% if post.has_previous %}
        <a href="?page={{ post.previous_page_number }}">previous</a>
    {% endif %}

    <span class="current">
            Page {{ post.number }} of {{ contacts.paginator.num_pages }}.
        </span>

        {% if post.has_next %}
        <a href="?page={{ post.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>

and my view looks like this;

# Main page
def index(request):
list_of_posts = Post.objects.all().order_by('-date')
list_of_posts = list_of_posts.filter(published=True)
paginator = Paginator(list_of_posts, 10)
page = request.GET.get('page')
try:
    post = paginator.page(page)
except PageNotAnInteger:
    # If page is not an integer, deliver first page.
    post = paginator.page(1)
except EmptyPage:
    # If page is out of range (e.g. 9999), deliver last page of results.
    post = paginator.page(paginator.num_pages)

return render_to_response('index.html', {'list_of_posts': list_of_posts})

I get a stange feeling the TypeError is due to the paginator not outputting any value? My traceback isnt too helpful but here it is

Exception Value:    
int() argument must be a string or a number, not 'NoneType'
Exception Location: C:\Python27\lib\site-packages\django\core\paginator.py in        validate_number, line 23
TypeError at /
int() argument must be a string or a number, not 'NoneType'

Any guidance on what could be going wrong would be much appreciated.

like image 242
jayduff2 Avatar asked Jan 23 '26 09:01

jayduff2


1 Answers

Add TypeError in your first catch:

except (PageNotAnInteger, TypeError):
    # ...

But you also could avoid this error if you will get page number like this:

page = request.GET.get("page", 1)
like image 104
Rost Avatar answered Jan 24 '26 22:01

Rost



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