Question

我需要使用序列化程序序列化数据，并且我还有一个用于保存的文件，但我无法将带有文件的额外变量路径到我的序列化程序

issue_dict = request.data.get('issue')
file = request.data.get('file')

这很好用：

 serializer = WriteIssueSerializer(data=issue_dict, context=self.get_serializer_context())

这就是我想要的，但它说＆＃34;得到了一个意想不到的关键字参数＆＃39; file＆＃39;＆＃34; ：

 serializer = WriteIssueSerializer(data=issue_dict, file=file, context=self.get_serializer_context())

我明白在序列化程序中我应该定义变量＆＃34; file＆＃34;，所以看看这个序列化器：

class WriteIssueSerializer(serializers.ModelSerializer):
    notes = IssueNoteSerializer(many=True)

    def create(self, val):
        issue_dict = val.get('issue')
        # issue_dict['assigned_to'] = issue_dict['assigned_to']['id']
        # issue_dict['reported_by'] = issue_dict['reported_by']['id']

        assigned_to_id = issue_dict.pop('assigned_to').id
        reported_by_id = issue_dict.pop('reported_by').id
        notes_info = issue_dict.pop('notes')
        # print(validated_data.pop('file'))
        issue = Issue.objects.create(assigned_to_id=assigned_to_id, reported_by_id=reported_by_id, **issue_dict)
        for note_info in notes_info:
            note = IssueNote.objects.create(**note_info)
            note.issue = issue

很明显，改变了 def create(self, val):到def create(self, val, file):会修复我的错误，但不，错误仍然相同

Answer 1

serializer = WriteIssueSerializer(data=issue_dict, file=file, context=self.get_serializer_context())

这会调用WriteIssueSerializer（__init__()）的构造函数，而不是.create()。所以你必须在那里创建额外的参数，或者调用.create()。

无法为Python（Django）函数调用添加额外的参数

1 个答案: