Mysql 如何在Django Views.py中使用distinct?
Views.pyMysql 如何在Django Views.py中使用distinct?,mysql,django,Mysql,Django,Views.py def user_detail(request): user_id = request.session.get('user_id') user = User.objects.get(username = request.user.get_username()) submit = Submit() submit_list = Submit.objects.filter(author=user).filter(result='AC').order_by('problem').dis
def user_detail(request):
user_id = request.session.get('user_id')
user = User.objects.get(username = request.user.get_username())
submit = Submit()
submit_list = Submit.objects.filter(author=user).filter(result='AC').order_by('problem').distinct()
context = {'User': User, 'submits':submit_list}
return render(request, 'koj/user_detail.html', context)
detail.html
Solved :
{% for submit in submits%}
<th> {{ submit.problem }} <th>
{% endfor %}
我想要的=>
Name : ziho2463
Solved : 1000 1001 1002
这是我的密码。筛选器工作正常,但当我添加“.distinct('problem')”时,出现“此数据库后端不支持字段上的distinct”错误
我还使用了.values('problem').distinct('problem')或.values_list('problem',flat=True).distinct('problem'))
但那个代码并没有给我想要的结果
=>
我怎样才能解决这个问题?你可以试试这个
Submit.objects.filter(author=user, result=AC).order_by('problem').values('problem').distinct()
Submit.objects.filter(author=user,result=AC)。order_by('problem')。values('problem')。distinct()
我以前尝试过,但出现了“此数据库后端不支持字段的distinct ON”错误
Name : ziho2463
Solved : 1 2 3
Submit.objects.filter(author=user, result=AC).order_by('problem').values('problem').distinct()