if same id is present in two tables show edit button else show assign button in django - django-models

Question

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.
Views.py
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})
templates
{% for list in list %}
{% if obj != list.id %}
<td>
<a href="/view_client_process/{{ list.id }}"><button
class="btn btn-info">Edit</button></a>
</td>
{% else %}
<h6>welcome</h6>
<td>
<a href="/view_client_process/{{ list.id }}"><button
class="btn btn-info">Assign</button></a>
</td>
{% endif %}
{% endfor %}

Answer 1

You can construct a set of username_ids and pass this to your template:
def Employee(request):
empS = Emp_Profile.objects.filter(is_active=True)
empprocess = set(Emp_Process.objects.values_list('username_id', flat=True).distinct())
return render(request, 'employee.html', {'emps' : emps, 'empprocess': empprocess })
In the template, we can then make a membership check of the set:
{% for emp in emps %}
<td>
{% if emp.id not in empprocess %}
<button class="btn btn-info">Edit</button>
{% else %}
<button class="btn btn-info">Assign</button>
{% endif %}
</td>
{% endfor %}
Note: you might want to rename your field username to user since a ForeignKey to a user is not the same as a username.
Note: please use {% url ... %} template tags [Django-doc] instead of performing URL processing yourself.