Aam*_*amu 17 django django-rest-framework
我使用Django作为后端服务器,使用Vue.js作为前端Movie应用程序.
我有一个Ticket模型
class MovieTicket(models.Model):
show = models.ForeignKey(Show)
seat = models.ForeignKey(Seat)
user = models.ForeignKey(User)
purchased_at = models.DateTimeField(default=timezone.now)
qrcode = models.ImageField(upload_to='qrcode', blank=True, null=True)
qrcode_data = models.CharField(max_length=255, unique=True, blank=True)
class Meta:
unique_together = ('show', 'seat')
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和它相关的Serializer
class MovieTicketSerializer(serializers.ModelSerializer):
class Meta:
model = MovieTicket
fields = '__all__'
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要购买新的Ticket,有一个视图映射到此URL http://dev.site.com/api/movies/buy-ticket/:
@api_view(['POST'])
@permission_classes([IsAuthenticated])
def buy_ticket(request):
serialized = MovieTicketSerializer(data=request.data)
if serialized.is_valid():
serialized.save()
return Response(serialized.data, status=status.HTTP_201_CREATED)
return Response(serialized._errors, status=status.HTTP_400_BAD_REQUEST)
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现在从前端(Vue.js)我可以创建一个新的电影票:
const formBody = {
show: this.$store.state.showSelected.showTime.id,
user: this.$store.state.user.id,
// selectedSeats is an array of seats that have been selected by the user. Here I am passing the first seat object.
seat: this.$store.state.selectedSeats[0].seat.id
};
this.$http.post("http://dev.site.com/api/movies/buy-ticket/", formBody)
.then(function (response) {
console.log(response.data);
})
.catch(function (response) {
console.log(response);
});
return;
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如果表单有效,这将创建一个新的MovieTicket对象,否则显示错误/ s.
现在,假设用户选择了多个席位,我可以遍历每个selectedSeats阵列并获得客户端的座位ID.并发布这样的东西:
{
"purchased_at": null,
"qrcode": null,
"qrcode_data": "",
"show": 11,
"seat": [
106,
219
],
"user": 34
}
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但令我困惑的是,seat.id如果Django休息框架只接受每个请求一个席位并相应地显示错误,我怎么能传递多个?如果故障单可用,则显示错误,以及是否为该show-seat创建可用的电影票证.
fet*_*zig 14
使用many = True初始化序列化程序
在您的实现中,这很容易实现:
serialized = MovieTicketSerializer(data=request.data, many=True)
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数据不是单个对象,而是一个对象数组.
您的信息提示您需要转换request.data以生成这些多个对象(所有相同的数据只是不同的座位号).对?
无论如何:
请参阅:如何使用Django Rest Framework创建多个模型实例?
编辑:
这里是drf文档中的信息:http://www.django-rest-framework.org/api-guide/serializers/#dealing-with-multiple-objects
(强烈建议在编写第一个真正的实现之前,从上到下阅读drf文档并只是玩它.有很多方法可以使用drf,并且知道所有这些都会导致更好的决策)
编辑2(问题更新后):
您可以从客户端发送此JSON(请参阅下文),或者buy_ticket(request)在调用之前从客户端在您的方法中发送的当前JSON创建此格式MovieTicketSerializer(...,many=True):
[
{
"purchased_at": null,
"qrcode": null,
"qrcode_data": "",
"show": 11,
"seat": 106,
"user": 34
},
{
"purchased_at": null,
"qrcode": null,
"qrcode_data": "",
"show": 11,
"seat": 219,
"user": 34
}
]
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这个答案是这个问题的一个很好的解决方案:
你可以简单地覆盖get_serializer法在APIView并传递many=True到get_serializer像这样的基础视角:
class SomeAPIView(CreateAPIView):
queryset = SomeModel.objects.all()
serializer_class = SomeSerializer
def get_serializer(self, instance=None, data=None, many=False, partial=False):
if data is not None:
data.is_valid(raise_exception=True)
return super(SomeAPIView, self).get_serializer(instance=instance, data=data, many=True, partial=partial)
else:
return super(SomeAPIView, self).get_serializer(instance=instance, many=True, partial=partial)
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如原始帖子评论中所述,您还必须data.is_valid()在将data关键字传递给序列化程序的情况下调用。
您可以在查看功能中查看座位数并创建一张或多张票:
@api_view(['POST'])
@permission_classes([IsAuthenticated])
def buy_ticket(request):
# Check if seats is a list
if isinstance(request.data['seat'], list):
seats = request.data.pop('seat')
models = []
for seat in seats:
# validate each model with one seat at a time
request.data['seat'] = seat
serializer = MovieTicketSerializer(data=request.data)
serializer.is_valid(raise_exception=True)
models.append(serializer)
# Save it only after all seats are valid.
# To avoid situations when one seat has wrong id
# And you already save previous
saved_models = [model.save() for model in models]
result_serializer = MovieTicketSerializer(saved_models, many=True)
# Return list of tickets
return Response(result_serializer.data)
# Save ticket as usual
serializer = MovieTicketSerializer(data=request.data)
serializer.is_valid(raise_exception=True)
serializer.save()
return Response(serializer.data)
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它会起作用,但老实说,这真是一团糟。您可以在不同的功能中移动用于创建座位的逻辑,使其看起来更好。
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