1

我有这个加入：

lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name)

这是网站

如果您尝试姓氏：Abbas 和名字：Amr，它会告诉您 amr abbas 有 1 个同学。

但是，如果您仅尝试 First name，则表明数据库中没有名为 amr 的律师（显然有）。

如果我更改(last__iexact=last_name)为，(last__icontains=last_name)则将姓氏留空可以正常工作并找到 amr。

但是last__icontains=last_name如果你搜索“collin”，你也会得到“collins”和“collingwood”，这不是我想要的。

你知道我可以如何使用iexact它，如果它是空白的，也可以忽略它吗？

谢谢

这是视图函数：

def search_form(request):
    if request.method == 'POST':
        search_form = SearchForm(request.POST)
        if search_form.is_valid():
            last_name = search_form.cleaned_data['last_name']
            first_name = search_form.cleaned_data['first_name']
            lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name)
            if len(lawyers)==0:
                form = SearchForm()
                return render_to_response('not_in_database.html', {'last': last_name, 'first': first_name, 'form': form})
            if len(lawyers)>1:
                form = SearchForm(initial={'last_name': last_name})
                return render_to_response('more_than_1_match.html', {'lawyers': lawyers, 'last': last_name, 'first': first_name, 'form': form}) 
            q_school = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('school', flat=True)
            q_year = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('year_graduated', flat=True)
            lawyers1 = Lawyer.objects.filter(school__iexact=q_school[0]).filter(year_graduated__icontains=q_year[0]).exclude(last__icontains=last_name)
            form = SearchForm()
            return render_to_response('search_results.html', {'lawyers': lawyers1, 'last': last_name, 'first': first_name, 'form': form})
    else:
        form = SearchForm()
        return render_to_response('search_form.html', {'form': form, })
4

1 回答 1

4

您不必一口气构建 QuerySet。

lawyers = Lawyer.objects.all()
if last_name:
    lawyers = lawyers.filter(last__iexact=last_name)
if first_name:
    lawyers = lawyers.filter(first__icontains=first_name)

Django 在需要之前不会评估 QuerySet（在这种情况下，len() 调用会强制它评估），因此您可以整天保持堆叠过滤器，直到您准备好运行查询。

http://docs.djangoproject.com/en/dev/ref/models/querysets/#when-querysets-are-evaluated

此外，您无需稍后创建新的 QuerySet，只需使用现有的即可。

q_school = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('school', flat=True)
q_year = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('year_graduated', flat=True)

可：

q_school = lawyers.values_list('school', flat=True)
q_year = lawyers.values_list('year_graduated', flat=True)
于 2010-01-20T17:43:42.120 回答