drupal-7 - 如何发送视图菜单路径作为参数

Question

 I define a menu in drupal7 that look something like:
       school
       --people
       ----teachers
       ----pupils

我创造：

1. 人们的内容类型。

2. 具有上述标题的菜单。

When I add person to people’s content, I choose if he belong to 'teachers' or 'pupils' in the menu.

Now I need that pressing on the menu will give me a list of all that kind of people. For example, pressing on ‘school->people->teachers’ will give me the list of all the teachers, one line for each teacher.

So I thing to build a view that get the menu/sub menu as parameter and give me the appropriate output.

But I don’t know how:

1. Send to a view the menu / sub menu as parameters. (How to send parameters to a view)

2. How to filter the view by this parameter and create the specific page.

Answer 1

创建 2 个视图页面（教师和学生）

您将在哪里显示所有教师或学生的名单

在菜单----教师中添加视图页面教师的路径。

在菜单----学生中添加视图页面学生的路径。

单击菜单后，您将重定向到所需的页面

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