Render html content stay on current page with form action set to different url in Django

My question is there is a way to render html content(bootstrap alert) on the current page(1st page) with POST action set to a different url(2nd page)?

I have following content in views.py:

def calendar(req): # 1st page
    return render(req,'calendar.html')

def result(req): # 2nd page
    if req.method == 'POST':
        xxx
        else:
            no_data = "xxx"
            context = { 'no_data': no_data }
            return render(req, 'calendar.html', context)
        xxx
        return render(req,'result.html',context)

The form info is submitted to different url(2nd page) in calendar.html:

{% if no_data %}  
    <div class="alert alert-warning" role="alert">
        {{ no_data }}
    </div> 

<form action="{% url 'result' %}" method="POST">

Thank you.

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