Question

我需要做的是，我需要转到一个列表视图到另一个列表视图。但是我需要使用单击的链接来过滤后面的列表视图。那么filter属性是什么？

我有一个带有批次（sbtc）的学生模型，我需要使用选定批次的过滤器传递稍后的列表视图。

@method_decorator(login_required, name='dispatch')
class FindStudent(ListView):
    template_name = 'Dashboard/findStudent.html'
    model = Student
    fields = ['sbtc']

    def get_queryset(self):
        batch = Student.objects.values_list('sbtc').distinct()
        return batch

@method_decorator(login_required, name='dispatch')
class FindStudentdetail(ListView):
    template_name = 'Dashboard/findStudentdetail.html'
    model = Student
    fields = ['all']

    def get_queryset(self):
        student = Student.objects.filter(sbtc=#here what will be)
        return student

Answer 1

在模板中，您只需添加指向findStudentDetail模板的链接即可；这是docs的修改示例：

<h1>Articles</h1>
<ul>
{% for article in object_list %}
    <li>{{ article.pub_date|date }} - {{ article.headline }}</li>\
    <a href="/articleDetail.html?smbc={{article.smbc}}">View Detail</a>
{% empty %}
    <li>No articles yet.</li>
{% endfor %}
</ul>

然后，在响应视图中，您只需使用kwargs即可获取GET参数：

class FindStudentdetail(ListView):
    template_name = 'Dashboard/findStudentdetail.html'
    model = Student
    fields = ['all']

    def get_queryset(self):
        student = Student.objects.filter(sbtc=self.kwargs['sbtc'])
        return student

Answer 2

def get_queryset(self):
    student = Student.objects.filter(sbtc=self.kwargs['pk'])
    return student

如何在Django中链接到页面

2 个答案: