Models: class Technology(models.Model): name = models.CharField(max_length=100, unique=True) slug = models.SlugField(max_length=100, unique=True)
class Site(models.Model): name = models.CharField(max_length=100, unique=True) slug = models.SlugField(max_length=100, unique=True) technology = models.ManyToManyField(Technology, blank=True, null=True) Views: def portfolio(request, page=1): sites_list = Site.objects.select_related('technology').only('technology__name', 'name', 'slug',) return render_to_response('portfolio.html', {'sites':sites_list,}, context_instance=RequestContext(request)) Template: {% for site in sites %} <div> {{ site.name }}, {% for tech in site.technology.all %} {{ tech.name }} {% endfor %} </div> {% endfor %} But in that example each site makes 1 additional query to get technology list. Is there any way to make it in 1 query somehow? -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.