我使用Django标记来调用url来调用html页面,其中我创建了一个按钮。但是,当我在urls.py中定义我的url并重新启动服务器时,我有一个internal server error。这是来自urls.py的代码
url(r'^layer_quarantine/', TemplateView.as_wiew(template_name='layer_quarantine.html'), name='layer_quarantine'),这里是我如何在我的over html文件中调用它。
<div class="page-header">
{% user_can_add_resource_base as add_tag %}
{% if add_tag %}
<a href="{% url "layer_upload" %}" class="btn btn-primary pull-right">{% trans "Upload Layers" %}</a>
<a style="margin-right:30px" href="{% url "layer_quarantine" %} class="btn btn-warning pull-right"></a>
{% endif %}
<h2 class="page-title">{% trans "Explore Layers" %}</h2>
</div>发布于 2019-08-28 07:04:25
您的urls中有一个错误:
url(r'^layer_quarantine/', TemplateView.as_view(template_name='layer_quarantine.html'), name='layer_quarantine'),https://stackoverflow.com/questions/57686554
复制相似问题