Question

在Django中，我希望能够上传.zip文件。该zip文件包含我要阅读和处理的scl文件（基本上是纯文本）。我要保存在本地的.zip和.scl。

问题是我无法通过upload_to方法将本地提取的.scl文件保存到模型中。在上传.zip和提取scl文件后，它会出现在项目的根目录中。我当前的解决方案如下，此处提取了zip，然后将其移动，但是这些路径是硬编码的，如果两个文件具有相同的名称，则会引发错误。如果找到了相似的命名文件，upload_to方法将重命名文件，并将重命名的文件路径保存到数据库。

models.py

class Default_software(models.Model):
    version = models.CharField(max_length=10, help_text='AISA_V2-Q')
    zip_location = models.FileField(upload_to='default_software') #docfile
    scl_location = models.FileField(upload_to='default_scl')
    created_at = models.DateTimeField(auto_now_add=True)

    def __str__(self):
        """String for representing the Model object"""
        return self.name

views.py

from django.views.decorators.csrf import csrf_exempt

from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.urls import reverse

from myapp.models import Default_software
from myapp.forms import DocumentForm

# import aboslute URL for files processing
from django.conf import settings

# file handling and saving
from shutil import move

# creating django file objects
from django.core.files import File

# zipprocessing
import zipfile

@csrf_exempt
def list(request):
    # Handle file upload
    if request.method == 'POST':
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            print(request.FILES)
            process_file(request.FILES['docfile'])

            # Redirect to the document list after POST
            return HttpResponseRedirect(reverse('list'))
    else:
        form = DocumentForm() # A empty, unbound form

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


    # Render list page with the documents and the form
    return render(request, 'list.html', {'default_software_list': software_list, 'form': form})


def process_file(upload):
    zip = zipfile.ZipFile(upload)
    # extract zip and search for .scl file
    zip_contents = zip.namelist()
    scl = zip.extract(zip_contents[0]) # scl file
    print("scl type: ", type(scl))
    with open(scl) as f:
        for line in f:
            pass
            #print(line)
    # move scl file to correct directory
    move('test.SCL', 'media/default_scl')
    # Save all to database
    #newdoc = Default_software(zip_location=upload, scl_location='default_scl/test.SCL', version='dunno')
    newdoc.save() # save all contents

list.html

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
        <title>Minimal Django File Upload Example</title>
    </head>

    <body>
        <!-- List of uploaded documents -->
        {% if default_software_list %}
            <ul>
                {% for default in default_software_list %}
                    <li><a href="{{ default.zip_location.url }}">{{ default.zip_location.name }}</a>- {{ default.created_at }}</li>
                {% endfor %}
            </ul>
        {% else %}
            <p>No default software uploaded...</p>
        {% endif %}

        <!-- Upload form. Note enctype attribute! -->
        <form action="{% url "list" %}" method="post" enctype="multipart/form-data">
            {% csrf_token %}
            <p>{{ form.non_field_errors }}</p>

            <p>{{ form.docfile.label_tag }} {{ form.docfile.help_text }}</p>

            <p>
                {{ form.docfile.errors }}
                {{ form.docfile }}
            </p>

            <p><input type="submit" value="Upload"/></p>
        </form>
    </body>

</html>

zip保存到/ root_project / media / default_software

scl保存到/ root_project / media / default_scl

最后，我如何处理Django将提取的zip文件从文件根移动到模型指定路径中的压缩文件？

Django上传zip文件并将.zip和解压缩的文件保存到同一模型中的不同位置

0 个答案: