保存前编辑 django

本文介绍了保存前编辑 django-rest-framework 序列化程序对象的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想在保存之前编辑 django-rest-framework 序列化程序对象.这就是我目前的做法 -

I want to edit a django-rest-framwork serializer object before it is saved. This is how I currently do it -

def upload(request): if request.method == 'POST': form = ImageForm(request.POST, request.FILES) if form.is_valid(): # All validation rules pass obj = form.save(commit=False) obj.user_id = 15 obj.save()

如何使用 django-rest-framework 序列化器对象来实现?

How can I do it with a django-rest-framework serializer object?

@api_view(['POST','GET']) def upload_serializers(request): if request.method == 'POST': serializer = FilesSerializer(data=request.DATA, files=request.FILES) if serializer.is_valid(): serializer.save()

推荐答案

您可以在保存序列化程序之前编辑序列化程序的对象:

You can edit the serializer's object before save the serializer:

if serializer.is_valid(): serializer.object.user_id = 15 # <----- this line serializer.save()