Question

我的问题在于django orderby子句。我使用了两个queryset，一个是state_filter，另一个是party_filter。没有order by子句正确地附加值。使用order_by子句然后计数值不正确

我的观点是：

def const(request):
    states = Loksabha.objects.values('state_name').distinct('state_name')
    constituency=Loksabha.objects.values('constituency_name').distinct('constituency_name').filter(state_name='MAHARASHTRA')
    dataset = Loksabha.objects.all()
    state_filter=Loksabha.objects.filter(state_name='MAHARASHTRA',constituency_name='Kolhapur').order_by('id')


    party_filter=state_filter.values('party_name').annotate(Count('party_name'))
    crime_filter=childcrime_type.objects.filter(state_name='MAHARASHTRA')
    womencrime_fltr=womencrime_type.objects.filter(state='MAHARASHTRA')

    xdata=[]
    ydata=[]

    for b in party_filter:
        xdata.append(b['party_name'])
        ydata.append(b['party_name__count'])

没有order_by子句的答案是：

 [
    {'party': 'Shiv Sena', 'party__count': 2},
    {'party': 'Indian Nationlist Congress', 'party__count': 3},
    {'party': 'Nationlist Congress Party', 'party__count': 1},
    {'party': 'Republican Party of India(A)', 'party__count': 2},
    {'party': 'Bharatiya Janata Party', 'party__count': 1},
    {'party': 'Independent', 'party__count': 2}
]

带有order_by子句的

答案是：

[
   {'party': 'Shiv Sena', 'party__count': 1},
   {'party': 'Shiv Sena', 'party__count': 1},
   {'party': 'Indian Nationlist Congress', 'party__count': 1},
   {'party': 'Indian Nationlist Congress', 'party__count': 1},
   {'party': 'Indian Nationlist Congress', 'party__count': 1},
   {'party': 'Nationlist Congress Party', 'party__count': 1},
   {'party': 'Republican Party of India(A)', 'party__count': 1},
   {'party': 'Republican Party of India(A)', 'party__count': 1},
   {'party': 'Bharatiya Janata Party', 'party__count': 1},
   {'party': 'Independent', 'party__count': 1},
   {'party': 'Independent', 'party__count': 1}
]

请给我适当的解决方案

Answer 1

执行此操作时：

party_filter=state_filter.values('party_name').annotate(Count('party_name'))

它会执行aggregation。因此，您可以根据party_name

获得总计数

现在执行此操作时：

party_filter=state_filter.values('party_name').annotate(Count('party_name')).order_by('party_name')

它将执行group by。

注意：聚合用于完整的结果集，为单个（分组）行注释。

附加valuequeryset

1 个答案: