从Django中的数据库中删除元素

时间:2018-04-18 16:47:38

标签: python django django-2.0

我正在尝试开发一个能够使用CRUD文件的Django应用程序。 目前我已经开发了上传下载功能,但是当我想要删除项目时,事情变得非常困难。

* Views.py *

def list(request):
# Handle file upload
if request.method == 'POST':
    form = DocumentForm(request.POST, request.FILES)
    all_docs = Document.objects.all()
    repeated = False
    if form.is_valid():
        for doc in all_docs:
            if Document(docfile=request.FILES['docfile']).docfile == doc.docfile:
                repeated = True
        if not repeated:
            newdoc = Document(docfile=request.FILES['docfile'])
            newdoc.save()

            # Redirect to the document list after POST
            return HttpResponseRedirect(reverse('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(
    request,
    'list.html',
    {'documents': documents, 'form': form}
)

* Forms.py *

class DocumentForm(forms.Form):
docfile = forms.FileField(
    label='Select a file'
)

* List.html *

<body>
    <!-- List of uploaded documents -->
    {% if documents %}
        <ul>
            {% for document in documents %}
                <li><a href="{{ document.docfile.url }}">{{ document.docfile.name }}</a></li>
            {% endfor %}
        </ul>
    {% else %}
        <p>No documents.</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>

*输出*

enter image description here

正如您所看到的,当我上传文档时,它会存储在数据库中,但我不知道如何制作按钮以便能够选择项目并将其删除,就像我做的那样上传系统。

Document的模型有一个属性ID和一个FileField。

我知道这是一个非常具体和密集的问题,但我花了太多时间研究它并且我没有得到结果。我开始开发这段代码以使视图能够删除该项目,但我不知道如何选择它:

*删除查看*

def remove_document(request):

if request.method == 'POST':
    form = DocumentForm()
    # Dont know how to reference the item I want to delete
    document = request.POST.objects.get(docfile = ?????)

    db_documents = Document.objects.all()

    for db_doc in db_documents:
        if db_doc = document:
            document.delete()

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

我是Django的新手,它的架构模型对我来说非常棘手

谢谢!

0 个答案:

没有答案