My Own Like Button: Django + Ajax — How?

Question

So I\'ve been having trouble turning this view into an Ajax call:

def company_single(request, slug):
    company = get_object_or_404(CompanyProfile, slug=slu         


        

Answer 1

Writing this here since I don't have enough reputation to comment and edits have to be at least 6 characters. In new versions of Django, you need to pass the path to the view function or the name of the url to the url template tag as a string. Therefore line 7 of above template would be:

url: "{% url 'like' %}",

Here is the part of the documentation that backs this up.