javascript - 如何在没有jquery的情况下从ajax发送整数？

Question

我正在尝试将一个名为“petadid”的整数从此 js 发送到名为“petadlikeview”的 django View 。但看起来数据没有出现在 View 中。如果我在 View 中打印“petadid”，它会显示“无”。谁能告诉我出了什么问题吗？我是 ajax 和 django 新手。

这是我的js:

<script>
$(document).ready(function(argument) {
$.ajaxSetup({cache:false});
var element = document.getElementById("likes");
var petadid = $(this).attr("data-petadid");
element.addEventListener("click",function(){
  var req = new XMLHttpRequest();
  req.open("POST",'/petadlike/')
  req.onload = function(){
    console.log(req.responseText);
    console.log(petadid);
    var data = JSON.parse(req.responseText);
  }
  req.setRequestHeader("X-CSRFToken", '{{ csrf_token }}');
  var data = {
    'petadid':petadid,
  }
  req.send(data);
});
});
</script>

这是我的 django View :

def petadlikeview(request):
print("vau")
if request.method=="POST":
    print("vai")
    petadid = request.POST.get('petadid')
    print(petadid)
    petad = PetAd.objects.get(pk=petadid)
    like_count = petad.petadlikes_set.all().count()
    like, created = petadlikes.objects.get_or_create(user=request.user,petad=petad)
    print(created)
    if created is False:
        petadlikes.objects.get(user=request.user,petad=petad).delete()
        like_count -= 1
        liked = False
    else:
        like.save()
        like_count += 1
        liked = True
    dict = {'like_count':like_count}
    return JsonResponse(dict)
    return HttpResponse(str(like_count)+' people have liked this')
else:
    return HttpResponse('Bad Request')

Answer 1

XMLHttpRequest 不会像 jQuery 那样自动将对象转换为 POST 数据，您需要自己创建 URL 编码的字符串。

var data = "petadid=" + encodeURIComponent(petadid);

还有

var petadid = $(this).attr("data-petadid");

应该是

var petadid = $(element).attr("data-petadid");