Create dynamic action URL form in Django python-CodePudding

whenever I submit the form it takes me to the testformresult page like https://www.studnentdetail.com/student/testformresult but What I want is, when someone enters the name of the student in the form.html page then data of that student is fetched from the mysql database and displayed on another page something like https://www.studentdetail.com/student/<student_name>. Also I want to know how like many railway site when we search for particular train like 10029 then in the next page it shows url something like https://www.train/10029-running-status.

form.html

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

    <label for="fname">First Name</label>
    <input type="text" id="fname" name="firstname" placeholder="Your name..">

    <input type="submit" value="Submit">
  </form>

views.py

def testform(request):
    return render(request, 'testform.html')

def testformr(request):
    fname = request.POST.get('firstname')
    return render(request, 'testformresult.html')

urls.py

path('testform', views.testform, name='testform'),
path('student/testformresult/<slug:student>', views.testformresult, name='testformresult'),

CodePudding user response：

you can try it with a redirect: (and think about using a CSRF Token and enable CsrfViewMiddleware ... see django docs https://docs.djangoproject.com/en/4.0/ref/csrf/ !!)

form.html

<form method="POST">

    {% csrf_token %}    

    <label for="fname">First Name</label>
    <input type="text" id="fname" name="firstname" placeholder="Your name..">

    <input type="submit" value="Submit">
  </form>

views.py

def testform(request):
    if request.method == 'POST':
       ....
       fname = request.POST.get('firstname')
       return redirect('testformresult', student=fname)

    return render(request, 'testform.html')

urls.py

path('testform', views.testform, name='testform'),
path('student/testformresult/<slug:student>', views.testformresult, name='testformresult'),