Question

Models.py

class Season(models.Model):
     name = models.CharField()
     statuses = models.CharField()

class Match(models.Model):
     SHOT_CHOICES = (
     ('W', 'Win'),
     ('F', 'Fail'),
     ('D', 'Draw'),
     )
     season = models.ForeignKey(Season, on_delete=models.CASCADE)
     status = models.CharField(choices=STAT_CHOICE, max_length=1)

我用于创建匹配项的视图

class CreateMatches(generics.CreateApiView)
    def create(self, request, *args, **kwargs):
            serializer = self.get_serializer(data=request.data, many=True)
            serializer.is_valid(raise_exception=True)
             self.perform_create(serializer)
            headers = self.get_success_headers(serializer.data)

statuses是一个类似“ WFFWFDDW”的字符串，我需要从外键季节的状态中获取“ nth”个字符，将状态转换为“ nth”个匹配。

Answer 1

由于 status 模型的Match字段不依赖于请求中的任何输入数据，因此您可以通过覆盖 {{1} } 方法的 save() 模型。但是，您必须设置 Match

null=True

Answer 2

我通过重写perform_create函数来解决了该问题

def perform_create(self, serializer):
    if type(serializer.validated_data) == list:
        for iter,item in enumerate(serializer.validated_data):
            season = item['season']
            count = season.matches.count()
            status = season.statuses[count+iter]
            item.update({'status':status})
        serializer.save()

在Django Rest中创建多个对象时从外键表字段进行处理？

2 个答案: