Empresas
Empregos
  • Sobre nós
  • Soluções
    • Publicação de vagas
      Publique sua vaga e receba candidatos qualificados em 48h.
    • Avaliações de candidatos
      Mais de 500 testes técnicos e psicológicos, mais anti-fraude.
    • Headhunting
      Busca executiva personalizada do início ao fim.
    • Folha de Pagamento + EOR
      Dispersão de folha e EOR em mais de 15 países da LATAM.
  • Preços
  • Empregos

0

341
Visualizações
Ajax, Django: status 200 but throws error instead of success

I am trying to post comment with ajax, but it is not working. When I press button console is not printing any error (like 404 etc.), comment is not being posted. Object returned by error:

{readyState: 4, getResponseHeader: ƒ, getAllResponseHeaders: ƒ, setRequestHeader: ƒ, overrideMimeType: ƒ, …} abort: ƒ (e) always: ƒ () catch: ƒ (e) done: ƒ () fail: ƒ () getAllResponseHeaders: ƒ () getResponseHeader: ƒ (e) overrideMimeType: ƒ (e) pipe: ƒ () progress: ƒ () promise: ƒ (e) readyState: 4 responseText: "↵" setRequestHeader: ƒ (e,t) state: ƒ () status: 200 statusCode: ƒ (e) statusText: "OK" then: ƒ (t,n,r) proto: Object

In command line I see:

"POST / HTTP/1.1" 200 5572

When I change button to "submit" it is posting and responding with proper JSON like:

{"comment": {"id": 16, "author": 1, "content": "test", "post": 12}}

My code is below, any help is appreciated:

views.py

def homepage(request):
    profiles = Follow.objects.filter(follow_by=request.user.profile).values_list('follow_to', flat=True)
    posts = Post.objects.filter(author_id__in=profiles).order_by('-date_of_create')
    if request.method == 'POST':
        form = CommentForm(request.POST)
        if form.is_valid():
            pk = request.POST.get('pk')
            post = Post.objects.get(pk=pk)
            new_comment = Comment.objects.create(
                author = request.user.profile,
                post = post,
                content = form.cleaned_data['content']
            )
            return JsonResponse({'comment': model_to_dict(new_comment)}, status=200)
    form = CommentForm()
    context = {
        'posts': posts,
        'form': form
    }
    return render(request, 'posts/homepage.html', context=context)

template

<div class="comments" id="{{ post.pk }}" style="display: none">
            {% include 'posts/comments.html' %}
            <form action="" method="post" class="commentForm" data-url="{% url 'post_comments' post.pk %}">
                {% csrf_token %}
                <input type="hidden" name="pk" value="{{ post.pk }}">
                {{ form.as_p }}
                <button type="button" class="commentBtn" id="{{ post.pk }}">Comment</button>
            </form>

addComment.js

$(document).ready(function () {
    $('.commentBtn').click(function () {
        let serializedData = $('.commentForm').serialize();
        let btn = $(this);
        let id = btn.attr('id');
        console.log($(".commentForm").data('url'));
        $.ajax({
            url: $(".commentForm").data('url'),
            data: serializedData,
            type: 'post',
            dataType: 'json',
            success: function (data) {
                console.log(data);
                $(`#${id}.comments`).load('/posts/comments/' + data.post);
                $('textarea').val('');
            },
            error: function(textStatus) {
                console.log(textStatus)
            }
        })
    })
})

Edit: I used this question: Ajax request returns 200 OK, but an error event is fired instead of success , deleted dataType: 'json' and added contentType: 'application/json' and now I got 403 error.

over 4 years ago · Santiago Trujillo
1 Respostas
Responde à pergunta

0

As you have use $('.commentForm') to send data of form to backend it will get all forms with class commentForm and send data of all forms that's the reason its not working .Instead you can change this $('.commentForm').serialize() to $(this).closest(".commentForm").serialize().

Also , you are using #${id}.comments inside success function of ajax this will override any content inside #${id}.comments and load new content from url i.e : /posts/comments/.. .So , to avoid this one way would be surround your {% include 'posts/comments.html' %} with some outer div and then change your selector i.e : #${id}.comments > .yourouterdivclassname so new content will be loaded inside that div only thus form will not get removed .

over 4 years ago · Santiago Trujillo Relatório
Responde à pergunta
Encontrar trabalhos remotos

Descubra a nova forma de encontrar um emprego!

melhores empregos
Principais categorias de trabalho
Empresas
Postar vaga Preços Comercial
Jurídico
Termos e Condições Política de privacidade
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomende algumas ofertas para mim
Preciso de ajuda