Using slice filter, you can get a substring; then compare the substring with the 'A':
{% for TblCompanies in object_list %}
{% if TblCompanies.company_name|slice:':1' == 'A' %}
<tr class="customer-table">
<td>{{ TblCompanies.company_id }}</td>
<td>{{ TblCompanies.company_name }}</td>
<td>{{ TblCompanies.contact_phone }}</td>
<td>{{ TblCompanies.billing_address }}</td>
<td>{{ TblCompanies.contact_e_mail }}</td>
</tr>
{% endif %}
{% endfor %}
As @Matthias commented, it would be better to pass filtered object_list in view. Assuming object_list is a queryset object:
object_list = object_list.filter(company_name__startswith='A')
Sorintg
Sort the object_list before passing it to the template:
page = requests.REQUEST.get('page', 'A') # or Get from view parameter
# depending on url conf.
object_list = (object_list.filter(company_name__startswith=page)
.order_by('company_name'))
UPDATE
NOTE: Change app with actual app name.
urls.py:
url(r'^/path/to/site/customers/(?P<page>[A-Z])$', 'app.views.list_customers')
app/views.py:
from django.shortcuts import render
def list_compnaies(request, page):
object_list = (Customer.objects.filter(company_name__startswith=page)
.order_by('company_name'))
return render(request, 'customers/list.html', {
'object_list': object_list,
})
customers/list.html
{# Link to A .. Z customer pages %}
{% for page in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' %}
<a href="/path/to/site/customers/{{ page }}">{{ page }}</a>
{# Use {% url ... %} once you learn the url tag if possible to reduce duplicated hard-coded url #}
{% endif %}
{% for TblCompanies in object_list %}
<tr class="customer-table">
<td>{{ TblCompanies.company_id }}</td>
<td>{{ TblCompanies.company_name }}</td>
<td>{{ TblCompanies.contact_phone }}</td>
<td>{{ TblCompanies.billing_address }}</td>
<td>{{ TblCompanies.contact_e_mail }}</td>
</tr>
{% endfor %}
