一、文件上传
方法一：简单的文件上传
前端页面获取上传的文件后，后端通过request.FILES.get('file)来接收
打开一个新的文件，将接收的文件写入新文件中，读取文件时要使用file.chunks()方法
views.py
from devops.settings import BASE_DIR def file_add(request): filepath = os.path.join(BASE_DIR, 'data/scripts/') if request.method == 'POST': file = request.FILES.get('file') filename = file.name with open(filepath+filename,'wb') as f: for line in file.chunks(): f.write(line) return redirect(reverse('file_list')) return render(request, 'file_add.html')
urls.py
from django.urls import path from mfile import views urlpatterns=[ path('list/',views.filelist,name='file_list'), path('add/',views.file_add,name='file_add'), ]
templates/file_list.html
{% extends 'base.html' %}
{% block body%}
<form action="" method="post" class="form-horizontal" enctype="multipart/form-data">
{% csrf_token %}
<div>
<input type="file" name="file">
</div>
<button class="btn btn-primary">提交</button>
</form>
{% endblock %}
方法二：
通过ModelForm实现，将文件路径保存在数据库中
settings.py
# 配置文件保存的目录信息 FILE_ROOT = os.path.join(BASE_DIR,'data/scripts/')
urls.py
from django.urls import path from mfile import views urlpatterns = [ path('filelist/', views.FileList.as_view(), name='file_list'), path('fileadd/', views.FileAdd.as_view(), name='file_add'), ]
models.py
from django.db import models from devops import settings class File(models.Model): title = models.TextField(max_length=32) file = models.FileField(upload_to=settings.FILE_ROOT)
forms.py
from django import forms from mfile import models class FileForm(forms.ModelForm): class Meta: model = models.File fields = '__all__'
views.py
from django.views import View from mfile import models, forms class FileList(View): def get(self, request): files = models.File.objects.all() return render(request, 'file_list.html', {'files': files}) class FileAdd(View): def get(self,request): form_obj = forms.FileForm() return render(request,'file_add.html',{'form_obj':form_obj}) def post(self,request): form_obj = forms.FileForm(request.POST,request.FILES) if form_obj.is_valid(): form_obj.save() return redirect(reverse('file_list'))
templates/file_list.html
{% extends 'base.html' %}
{% block body %}
<table class="table">
<tr>
<td>序号</td>
<td>标题</td>
<td>文件路径</td>
</tr>
{% for file in files %}
<tr>
<td>{{ forloop.counter }}</td>
<td>{{ file.title }}</td>
<td>{{ file.file}}</td>
</tr>
{% endfor %}
</table>
<div class="form-group">
<button>
<a href="{% url 'file_add' %}" >文件上传</a>
</button>
</div>
{% endblock %}
templates/file_add.html
{% extends 'base.html' %}
{% block body%}
<form action="" method="post" class="form-horizontal" enctype="multipart/form-data">
{% csrf_token %}
<div>
<span>标题</span>
<input type="text" name="title">
</div>
<div>
<input type="file" name="file">
</div>
<button class="btn btn-primary">提交</button>
</form>
{% endblock %}
在数据库表中保存了文件信息和保存的路径
在访问filelist时可显示文件信息
二、文件下载
前端页面提供一个下载文件的id，views通过文件id从数据库中查出文件保存路径，将文件返回给用户即可。
urls.py
from django.urls import path from mfile import views urlpatterns = [ path('filelist/', views.FileList.as_view(), name='file_list'), path('fileupload/', views.FileUpload.as_view(), name='file_upload'), path('filedownload/<int:id>/', views.FileDownload.as_view(), name='file_download'), ]
views.py
from django.views import View from devops.settings import FILE_ROOT from mfile import models, forms class FileDownload(View): def get(self,request,id): file_obj = models.File.objects.get(id=id).file # 获取文件名称 filename = str(file_obj).split('/')[-1] filepath = os.path.join(FILE_ROOT,filename) #这里不能用with open打开文件 file = open(filepath,'rb') rep = FileResponse(file) rep['Content-Type'] = 'application/octet-stream' rep['Content-Disposition'] = 'attachment;filename="%s"'%filename return rep
templates/filelist.html
{% extends 'base.html' %}
{% block body %}
<table class="table">
<tr>
<td>序号</td>
<td>标题</td>
<td>文件路径</td>
</tr>
{% for file in files %}
<tr>
<td>{{ forloop.counter }}</td>
<td>{{ file.title }}</td>
<td>{{ file.file}}</td>
<td><a href="{% url 'file_download' file.id %}"><button class="btn btn-primary">下载</button></a></td> #给后端传一个id值
</tr>
{% endfor %}
</table>
<div class="form-group">
<button>
<a href="{% url 'file_upload'%}" >文件上传</a>
</button>
</div>
{% endblock %}
返回文件有三种方式：
方式一：使用HttpResponse
from django.shortcuts import HttpResponse class FileDownload(View): def get(self,request,id): file_obj = models.File.objects.get(id=id).file # 获取文件名称 filename = str(file_obj).split('/')[-1] filepath = os.path.join(FILE_ROOT,filename) #这里不能用with open打开文件 file = open(filepath,'rb') rep = HttpResponse(file) rep['Content-Type'] = 'application/octet-stream' rep['Content-Disposition'] = 'attachment;filename="%s"'%filename return rep
方式二：使用StreamingHttpResponse
from django.http import StreamingHttpResponse class FileDownload(View): def get(self,request,id): file_obj = models.File.objects.get(id=id).file # 获取文件名称 filename = str(file_obj).split('/')[-1] filepath = os.path.join(FILE_ROOT,filename) #这里不能用with open打开文件 file = open(filepath,'rb') rep = StreamingHttpResponse(file) rep['Content-Type'] = 'application/octet-stream' rep['Content-Disposition'] = 'attachment;filename="%s"'%filename return rep
方式三：使用FileResponse（推荐）
from django.http import FileResponse class FileDownload(View): def get(self,request,id): file_obj = models.File.objects.get(id=id).file # 获取文件名称 filename = str(file_obj).split('/')[-1] filepath = os.path.join(FILE_ROOT,filename) #这里不能用with open打开文件 file = open(filepath,'rb') rep = FileResponse(file) rep['Content-Type'] = 'application/octet-stream' rep['Content-Disposition'] = 'attachment;filename="%s"'%filename return rep
三种http响应对象在django官网都有介绍。https://docs.djangoproject.com/en/1.11/ref/request-response/
推荐使用FileResponse，从源码中可以看出FileResponse是StreamingHttpResponse的子类，内部使用迭代器进行数据流传输。