Django将ajax用于自定义django like按钮

Question

So I am trying to create my own custom like button with ajax but there is one problem, when I press the like button I should reload the page in order to see my upvote. I want to do it without reloading the page.

here is the code:

article.html

{% block content %}
<div>
    {% for a in article %}
    [... unrelated html ...]

    <p><input type="button" class="like" id="{{ a.id }}" value="Like" /></p>
    <p id="count{{ a.id }}">{{ a.total_likes }}</p>
    </div>
    {% endfor %}

</div>
<script>

$('.like').click(function(){

      $.ajax({
               type: "POST",
               url: "{% url 'like_button' %}",
               data: {'pk': $(this).attr('id'), 'csrfmiddlewaretoken': '{{ csrf_token }}'},
               dataType: "json",
               success: function(response) {
                      var pk = $(this).attr('id');
                      $('#count' + pk).html(response.likes_count) #change the html when success. response.likes_count is connected to python, it indicates the number of counts on the query.
                },
               error: function(rs, e) {
                      alert(rs.responseText);
               }
          });
    })

</script>
{% endblock %}

What should I do in order to reslove this issue?

Answer 1

Inside success callback this doesn't point to DOM element

$('.like').click(function(){
  var pk = $(this).attr('id'); // Get your id here
  // Whole ajax stuff
})