django 1.5如何从内存中读取csv

Sim*_*eth 5 csv django file-upload

我试图找出如何读取上传的CSV而不将其保存到磁盘...

我被困在form.cleaned_data ['file'].读...我似乎没有得到任何输出

如果我只能弄清楚如何获得输出,那么我可以编写一个适当的函数来处理数据行.

#addreport.html
<form enctype="multipart/form-data" method="post" action="/products/addreport/">   {%csrf_token %}
<table>
{{ form.as_table }}
</table>
<input type="submit" value="Submit" />
Run Code Online (Sandbox Code Playgroud)

#forms.py
from django import forms
#
class UploadFileForm(forms.Form):
    file  = forms.FileField()
Run Code Online (Sandbox Code Playgroud)

-

# views.py
def addreport(request):
if request and request.method == "POST":
    form = UploadFileForm(request.POST, request.FILES)
    if form.is_valid():
               print form.cleaned_data['file'].read()
    else:
        print form.errors
        print request.FILES
        #form = UploadFileForm()
else:
    form = UploadFileForm()

return render_to_response('products/addreport.html', {'form': form},context_instance=RequestContext(request))
Run Code Online (Sandbox Code Playgroud)

Hie*_*yen 6

我的一个项目的设置非常类似.你的表格有干净的方法吗?

否则我认为你可以从中读取文件form.cleaned_data['file']:

import csv

def addreport(request):
    if request.method == "POST":
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            reader = csv.reader(form.cleaned_data['file'])
            for row in reader:
                print row
        else:
            print form.errors
            print request.FILES
            #form = UploadFileForm()
    else:
        form = UploadFileForm()

    return render(request, 'products/addreport.html', {'form': form})
Run Code Online (Sandbox Code Playgroud)

如果它不起作用,请尝试request.FILES['file']直接使用而不是form.cleaned_data['file'].

希望能帮助到你.