首先上结构
mynode -> app5 -> urls.py & views.py
| -> templates -> 5 -> upload.html
| -> mynode -> urls.py
| -> media
按照顺序,先上app5/urls.py- from django.urls import path<br>from app5 import views as v5<br><br>app_name = 'app5'<br><br>urlpatterns = [<br> path('upload_file/', v5.upload_file, name = 'upload_file'),<br> path('show_upload/', v5.show_upload, name = 'show_upload'),<br>]
复制代码 path('upload_file/', v5.upload_file, name = 'upload_file'),指定upload_file跳转功能
path('show_upload/', v5.show_upload, name = 'show_upload'),指定show_upload跳转功能
接着是app5/view.py
- from django.shortcuts import render<br>from django.http import HttpResponse
- import os
- def show_upload(request):
- return render(request, '5/upload.html')
- def upload_file(request):if request.method == 'POST':
- get_file = request.FILES.get('myfile',None)
- if get_file:
- path = 'media/uploads'
- if not os.path.exists(path):
- os.makedirs(path)
- dest = open(os.path.join(path,get_file.name),'wb+')
- for chunk in get_file:
- dest.write(chunk)
- dest.close()
- return HttpResponse('上传文件成功!')
- else:
- return HttpResponse('没有上传文件!')
复制代码
首先写了一个show_upload方法,跳转到初始页面
接下来是upload_file方法,首先判断请求方式是否是POST,接下来获取上传文件,指定上传路径,如果路径不存在就创建一个,把上传文件内容写到指定路径下
再来是templates/5/upload.html
- //这个是错误的<br><form enctype="multipart/form-data" action="{% url 'app5:upload_file' %}" method="post">
- {% csrf_token %}
- <input type="file" name="myfile" />
- <br/>
- <input type="submit" value="upload_file" />
- </form><br>
复制代码
指定了一个action,{% url 'app5:upload_file' %},app5是app5/urls.py中的app_name,upload_file则是要跳转连接,同时因为url已经指定这个连接要跳转的views中的功能,因此这个就是app5/view.py里面的upload_file方法
这个页面展示是正常的,但是在写好功能以后,无论怎么点提交,都没法跳转到upload_file功能
仔细看表单的名称 |