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django 1.5 如何从内存中读取 csv

转载 作者:行者123 更新时间:2023-12-04 20:10:26 25 4
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我试图找出如何在不将其保存到磁盘的情况下读取上传的 CSV ...

我被困在 form.cleaned_data['file'].read ......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合适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" />


#forms.py
from django import forms
#
class UploadFileForm(forms.Form):
file = forms.FileField()

——
# 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))

最佳答案

我的一个项目有非常相似的设置。你的表格中有任何干净的方法吗?

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

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})

如果它不起作用,请尝试使用 request.FILES['file']直接代替 form.cleaned_data['file'] .

希望能帮助到你。

关于django 1.5 如何从内存中读取 csv,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18159090/

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