title: if same id is present in two tables show edit button else show assign button in django
body: <p>Get all the employee profile  table id and check the id with employee process,if id matches show edit button in templates else show assign button.</p>
  
  <p>Views.py</p>
  
  <pre><code>def Employee(request):
      emp = Emp_Profile.objects.filter(is_active=True)
      emptable = Emp_Profile.objects.values_list('id')
      print(emptable)
      empprocess = Emp_Process.objects.values_list('username_id').distinct()
      print(empprocess)
      obj = {}
      for i in range(len(empprocess)):
          obj[i] = empprocess[i]
  
      return render(request, 'employee.html',{'list' : emp,'empprocess':empprocess,'obj':obj})
  </code></pre>
  
  <p>templates</p>
  
  <pre><code>{% for list in list %}
  {% if  obj != list.id %}
  &lt;td&gt;
          &lt;a href="/view_client_process/{{ list.id }}"&gt;&lt;button
                  class="btn btn-info"&gt;Edit&lt;/button&gt;&lt;/a&gt;
      &lt;/td&gt;
  {% else %}
  &lt;h6&gt;welcome&lt;/h6&gt;
  &lt;td&gt;
          &lt;a href="/view_client_process/{{ list.id }}"&gt;&lt;button
                  class="btn btn-info"&gt;Assign&lt;/button&gt;&lt;/a&gt;
      &lt;/td&gt;
  {% endif %}
  {% endfor %}
  </code></pre>