Question

我正在尝试上传csv文件，然后解析为xml格式但在保存之前＆＃34;导入csv文件并将其保存为xml文件＆＃34; 我有上传功能正常工作并正确保存;但是，我的问题是如何将上传的文件传递给解析为xml格式的函数。我正在使用django python

Model.py

from django.db import models
from django.urls import reverse
from .validators import validate_file_extension

# Create your models here.
class DataFile(models.Model):
    title = models.CharField(max_length=250)
    description = models.CharField(max_length=1000)
    file = models.FileField()

def __str__(self):
    return self.title + ' - ' +  self.description + ' - ' + self.file

这是我的form.py

from django import forms
from .models import DataFile


class FileForm(forms.ModelForm):
    class Meta:
        model = DataFile
        fields = ['title', 'description', 'file' ]

这是我的view.py

def importing(request):
    form = FileForm(request.POST or None)
    if request.method == 'POST':
        form = FileForm(request.POST, request.FILES)
        if form.is_valid():

            #here is where the passing supose to be 
            form = csv2xml(request.FILES['file']) 

            form.save()
            return HttpResponseRedirect('http://127.0.0.1:8000')

    return render(request, 'import.html', { 'form': form })

更新我希望将上传的文件传递给：

的函数

def csv2xml(csv_file):
    xml = dicttoxml.dicttoxml(csv_file)

    return xml

我希望将文件传递给csv2xml函数，进行转换，然后将其另存为xml

当我使用上述方法将其传递给函数时，它说：

AttributeError at /importing
'InMemoryUploadedFile' object has no attribute 'save'

Answer 1

试试这个答案

def importing(request):
    form = FileForm(request.POST or None)
    if request.method == 'POST':
        form = FileForm(request.POST, request.FILES)
        if form.is_valid():
            # here is where the passing supose to be 

            # Changes >>>>>>>>>
            form = csv2xml(request.FILES['file'].read())
            with open('myfile.xml', 'wb') as xml_file:
                xml_file.write(form)

            # form.save() No need

            # Changes <<<<<<<<<<<<
            return HttpResponseRedirect('http://127.0.0.1:8000')

    return render(request, 'import.html', {'form': form})

Answer 2

执行此操作的最佳方法是在save类的models.Model方法中进行一些更改。在models.py：

中有类似的内容

class DataFile(models.Model):
    title = models.CharField(max_length=250)
    description = models.CharField(max_length=1000)
    file = models.FileField()

    def save(self, *args, **kwargs):
        if self.file:
            """ Do Whatever you wanna do with the upload file here """
        super(DataFile, self).save(*args, **kwargs)

这将对上传的文件进行必要的更改，然后再保存在服务器上。

在将文件保存到django之前，将上传的文件解析为另一种格式

2 个答案: