Question

我正在尝试发送一个名为＆＃39; petadid＆＃39;的整数。从这个js到django视图叫做petadlikeview＆＃39;。但看起来数据并没有在视图中出现。如果我打印出了“petadid＆＃39;在视图中，它显示“无”。谁能告诉我有什么问题？我是ajax和django的新手。

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);

另外

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

应该是

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

如何在没有jquery的情况下从ajax发送一个整数？

1 个答案: