python - 在 Django 的引导模式中打开新视图

Question

我的情况是我有几个模式（确切地说是 4 个），应该可以在我的应用程序的每个视图中访问。我将仅使用一个作为示例。让我们看看我是如何在主屏幕视图上显示它的：

base.html

...abbreviated...

<a data-toggle="modal" href="#exampleModal">Shipment</buton>

<!-- Modal -->
<div class="modal fade" id="exampleModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
  <div class="modal-dialog" role="document">
    <div class="modal-content">
      <div class="modal-header">
        <h5 class="modal-title" id="exampleModalLabel">Create Shipment</h5>
        <button type="button" class="close" data-dismiss="modal" aria-label="Close">
          <span aria-hidden="true">&times;</span>
        </button>
      </div>
      <form method="POST">
        {% csrf_token %}
      <div class="modal-body">
        {{shipmentForm|crispy}}
      </div>

      <div class="modal-footer">
        <button type="button" class="btn btn-secondary" data-dismiss="modal">Close</button>
        <button type="submit" name="shipment" class="btn btn-primary">Save changes</button>
      </form>
      </div>
    </div>
  </div>
</div>

这是传递shipmentForm给该模板的视图

视图.py

def HomeView(request):
    if request.method == 'POST':
        shipment_form = CreateShipmentForm(request.POST)
        if shipment_form.is_valid():
            new_shipment = shipment_form.save()the user to the success page
            return redirect('ShipmentListView')

    shipmentForm = CreateShipmentForm()
    context = {
        'shipmentForm': shipmentForm,
    }
    return render(request, 'home.html', context)

这工作得很好，但问题是，我需要在我的应用程序的每个视图中包含来自 HomeView 的这段代码。那将是我自己的很多重复。我已经在我的 base.html 中包含了模态的 html，这样我就不必重复了。但不是我上面所说的，我想让启动模式的按钮看起来/做这样的事情。

base.html

<a data-toggle="modal" href="{% url 'CreateShipmentView' %}>Shipment</buton>

将模态表单的 html 放在单独的文件中：

模态的.html

<!-- Modal -->
<div class="modal fade" id="exampleModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
  <div class="modal-dialog" role="document">
    <div class="modal-content">
      <div class="modal-header">
        <h5 class="modal-title" id="exampleModalLabel">Create Shipment</h5>
        <button type="button" class="close" data-dismiss="modal" aria-label="Close">
          <span aria-hidden="true">&times;</span>
        </button>
      </div>
      <form method="POST">
        {% csrf_token %}
      <div class="modal-body">
        {{shipmentForm|crispy}}
      </div>

      <div class="modal-footer">
        <button type="button" class="btn btn-secondary" data-dismiss="modal">Close</button>
        <button type="submit" name="shipment" class="btn btn-primary">Save changes</button>
      </form>
      </div>
    </div>
  </div>
</div>

然后有这样的视图：

def CreateShipmentView(request):
    if request.method == 'POST':
        shipment_form = CreateShipmentForm(request.POST)
        if shipment_form.is_valid():
            new_shipment = shipment_form.save()the user to the success page
            return redirect('ShipmentListView')

    shipmentForm = CreateShipmentForm()
    context = {
        'shipmentForm': shipmentForm,
    }
    return render(request, 'modal.html', context)

如果不使用 django-modal-bootstrap 包，这怎么可能？我试过了，但它似乎非常敏感，我不喜欢它，所以宁愿避免这种情况。谁能指出我如何使这项工作的方向？

标签： pythondjangobootstrap-modal