django储存文件 Django接收照片储存文件的实例代码

时间:2021-04-21 PythonNew_Mr.Wang 人气:0

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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