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I'm creating a simple search application. I have a model with some data, and the index page is just a search bar that search results of that model. I'm creating the form using just HTML, not a proper Django Form.

index.html:

<form method="get" action="{% url 'core:search' %}?page=1&q={{ request.GET.q }}">
   <div class="form-floating mb-3">
      <input type="search" class="form-control" id="floatingInput" name="q" value="{{ request.GET.q }}">
      <label for="floatingInput">Type your search</label>
   </div>
   <button class="w-100 btn btn-lg btn-primary" type="submit">Search</button>
</form>

The results are paginated, so I need that the url requested by the form is http://127.0.0.1:8000/search/?page=1&q=query, query being the search term typed in the input. But what I wrote in the action parameter in the form doesn't work as I expected: even though it's writen action="{% url 'core:search' %}?page=1&q={{ request.GET.q }}, the URL requested is just http://127.0.0.1:8000/search/?q=query. The page param simply doesn't show up.

I wrote page=1 because, as is the search result, the first one requested is always the first page.

The view that process this request is the search_view. I'm putting it below just as more info, but I think the problem is my misunderstanding of url template tag in the action form parameter.

search_view:

def search_view(request):
    posts = Post.objects.all()
    query = request.GET.get('q')
    page_num = request.GET.get('page')

    paginator = Paginator(posts.annotate(
        search=SearchVector('post_title', 'post_subtitle'),
    ).filter(search=query).order_by('post_title'), 15)

    num_results = paginator.count
    num_pages = paginator.num_pages

    try:
        posts = paginator.page(page_num)
    except PageNotAnInteger:
        posts = paginator.page(1)
    except EmptyPage:
        posts = paginator.page(paginator.num_pages)

    context = {
        'posts': posts,
        'num_results': num_results,
        'num_pages': num_pages,
        'page_range': paginator.page_range,
    }

    return render(request, 'core/search.html', context)

How can I make the URL requested be like http://127.0.0.1:8000/search/?page=1&q=query ?

André Carvalho
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  • No need to send the page _always_, Just write: `request.GET.get('page', '1')`. Although aren't you handling that in your try except? Why do you need the page in the GET parameters for the first page anyway? – Abdul Aziz Barkat Sep 17 '21 at 03:35
  • Because I trying to create a paginator that have a previous and a next button, but it doesn't show all page numbers. For example, if I have 9 pages, I want that my paginator show only the active page (5, for example), two pages before that (3 and 4) and two after (6 and 7). For doing this I need to do some math with the page number. But as page is not passed in the first request, my code becomes not too DRY for my taste. – André Carvalho Sep 17 '21 at 04:02
  • `page.number`? For you since the page object is named `posts` you can write `posts.number`. See https://docs.djangoproject.com/en/3.2/ref/paginator/#django.core.paginator.Page.number – Abdul Aziz Barkat Sep 17 '21 at 05:25

0 Answers0