上传单个文件

使用一个模型来处理上传文件

node2:/django/mysite/news/templates#cat upload.html 
<div class="row">
<div class="col-md-8 col-md-offset-2">
<form class="form-inline" role="form"  method="post" enctype="multipart/form-data" accept-charset="utf-8">
<div class="form-group">
<input type="file" name="file">
</div>
<div class="form-group">
<input type="submit" value="上传文件">
</div>
</form>
</div>
</div>

node2:/django/mysite/mysite#cat urls.py |grep upload
url(r'^upload/$',newview.upload,name='upload'),

def upload(request):
    if request.method=="POST":
        print '1111111---------------------'
        print request.FILES['file']
        print '1111111---------------------'
        print '2222222---------------------'
        print str(request.FILES['file'])
        print '2222222---------------------'
        handle_upload_file(request.FILES['file'],str(request.FILES['file']))
        return HttpResponse('Successful') #此处简单返回一个成功的消息，在实际应用中可以返回到指定的页面中  
    else:
     return render_to_response('upload.html')

def handle_upload_file(file,filename):
    path='/tmp/'     #上传文件的保存路径，可以自己指定任意的路径  
    if not os.path.exists(path):
        os.makedirs(path)
    with open(path+filename,'wb+')as destination:
        for chunk in file.chunks():
            print '-----------'
            print chunk
            print '-----------'
            destination.write(chunk)