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Question

我正在用Need a minimal Django file upload example做一个复制猫

我用代码更改了 view.py，将 csv 文件转储到 sqlite 数据库中。 我已经在默认的 sqlite 数据库中创建了该表。

import sqlite3
import csv
import sys
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse

from myproject.myapp.models import Document
from myproject.myapp.forms import DocumentForm

def list(request):
    # Handle file upload
    if request.method == 'POST':
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            newdoc = Document(docfile = request.FILES['docfile'])
            newdoc.save()


    gfile= csv.reader(open(newdoc))
    gon = sqlite3.connect("database.sqlite")
    gon.text_factory = str
    gon.execute("DELETE FROM abc where rowID > 0 ")
    gon.executemany("insert into abc values (?, ?, ?, ?, ?)", gfile)
    gon.commit()
    gon.close()*

    return HttpResponseRedirect(reverse('myproject.myapp.views.list'))
    else:
        form = DocumentForm() # A empty, unbound form

    # Load documents for the list page
    documents = Document.objects.all()

    # Render list page with the documents and the form
    return render_to_response(
        'myapp/list.html',
        {'documents': documents, 'form': form},
        context_instance=RequestContext(request)
    )

我的代码从第一次引入 gfile 开始

第 20 行错误:强制转换为 unicode:需要字符串或缓冲区，找到文档 请帮忙

Answer 1

您正在将 Document 实例传递给 open。相反，您应该传递直接上传到 csv.reader 的文件:

gfile = csv.reader(request.FILES['docfile'])