Django接收照片储存文件的实例代码
(编辑：jimmy 日期: 2026/4/23 浏览：3 次 )
后端:
from rest_framework.views import APIView
from car import settings
from django.shortcuts import render, redirect, HttpResponse
from dal import models
from django.http import JsonResponse
import os
BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
class Image(APIView):
def post(self, request):
file_obj = request.FILES.get('send',None)
print("file_obj",file_obj.name)
file_path = os.path.join(BASE_DIR, 'media', 'user/img', file_obj.name)
print("file_path", file_path)
with open(file_path, 'w') as f:
for chunk in file_obj.chunks():
f.write(chunk)
message = {}
message['code'] = 200
return JsonResponse(message)
前端ajax:
<form method="post" action="/upload/" enctype="multipart/form-data" target="ifm1">
<input type="file" name="send"/>
<input type="submit" value="Form表单提交"/>
</form>
下面在看下在Django中接收文件并存储
首先是一个views函数的例子
def get_user_profiles(request):
if request.method == 'POST':
myFile = request.FILES.get("filename", None)
if myFile:
dir = os.path.join(os.path.join(BASE_DIR, 'static'),'profiles')
destination = open(os.path.join(dir, myFile.name),
'wb+')
for chunk in myFile.chunks():
destination.write(chunk)
destination.close()
return HttpResponse('ok')
这是一个简单的接收客户端上传的头像文件并保存的例子，应该看过这个就已经大体会使用接收文件了
但是这里的filename是客户端上传的文件名，也可能是像下面这样的表单
<input type="file" name="filename" />
如果不知道固定上传的文件名，想要客户端上传什么文件就以其上传的名字命名可以这么写
def get_user_profiles(request):
if request.method == 'POST':
if request.FILES:
myFile =None
for i in request.FILES:
myFile = request.FILES[i]
if myFile:
dir = os.path.join(os.path.join(BASE_DIR, 'static'),'profiles')
destination = open(os.path.join(dir, myFile.name),
'wb+')
for chunk in myFile.chunks():
destination.write(chunk)
destination.close()
return HttpResponse('ok')
不过这个是通过输出request.FILES试出来的，不知道是否有更合适的方法。
总结
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