Django_文件下载

 

一、小文件下载

1、视图 views.py

三种方式实现,任选其一

(1)使用HttpResponse

# 导入模块
from
django.shortcuts import HttpResponse def download(request):   file = open('crm/models.py', 'rb')   response = HttpResponse(file)   response['Content-Type'] = 'application/octet-stream' #设置头信息,告诉浏览器这是个文件   response['Content-Disposition'] = 'attachment;filename="models.py"'   return response

(2)使用StreamingHttpResponse

from django.http import StreamingHttpResponse
def download(request):
  file
=open('crm/models.py','rb')   response =StreamingHttpResponse(file)   response['Content-Type']='application/octet-stream'   response['Content-Disposition']='attachment;filename="models.py"'   return response

(3)使用FileResponse

from django.http import FileResponse
def download(request):
    file=open('crm/models.py','rb')
    response =FileResponse(file)
    response['Content-Type']='application/octet-stream'
    response['Content-Disposition']='attachment;filename="models.py"'
    return response

 

2、添加路由 urls.py

配置一个下载的路径

url(r'^download/',views.download,name="download"),

 

3、模板 templates 的修改

配置一个 a 标签,跳转地址配置要跳转的下载路径(对应的视图)

<div class="col-md-4"><a href="{% url 'download' %}" rel="external nofollow" >点我下载</a></div>

 

二、大文件下载

大文件需要使用迭代器优化

只需要修改 views.py 文件

from django.http import StreamingHttpResponse

def download(request):
  def file_iterator(file_name, chunk_size=512):
     with open(file_name, 'rb') as f:
        while True:
           c = f.read(chunk_size)
              if c:
                 yield c
               else:
                  break
   the_file_name = 'static/images/exam/logo.png'
   response = StreamingHttpResponse(file_iterator(the_file_name))
   response['Content-Type'] = 'application/octet-stream'
   response['Content-Disposition'] = 'attachment;filename="{0}"'.format(the_file_name)
   return response

 

 

 

posted @ 2019-12-25 12:05  Tester北柯郡  阅读(199)  评论(0编辑  收藏  举报