我正在尝试使用django中的下拉菜单,允许用户从菜单中选择一个选项,然后导航到页面。
具体而言。我正在建立一个小型项目管理系统作为一个学习项目(我知道我在这里有很多其他问题 - 我只是想让事情先发挥功能,然后担心风格并删除不需要的变量)。我希望用户从下拉菜单中选择一个项目(我可以填充),然后导航到显示项目详细信息的页面。我目前通过单击链接来实现此功能,但希望将其设置为用户可以选择现有项目并查看详细信息
html表单应该将用户从view_existing_projects视图带到view_project视图。现在我可以转换视图查找,但是project_id没有通过
urls.py
url(r'^view_project/(?P<project_id>\d+)/$', views.view_project, name='view_project'),
html表单
<form method="POST" action="/view_project/{{ project.id }}"/>{% csrf_token %}
<select name = "project_id">
{% for project in projects %}
<option value="{{ project.id }}" >{{ project.address1 }}</option>
{% endfor %}
</select>
<input type="submit" value="View Details" />
</form>
views.py
def view_existing_projects(request, user_id):
context = RequestContext(request)
user = User.objects.get(id=user_id)
projects = ProjectSite.objects.filter(owner__id=user.id)
args = {}
args.update(csrf(request))
args['users'] = user
args['projects'] = projects
if request.method == 'POST':
project_id = request.POST.get('project_id')
args['project_id']= project_id
return redirect('/view_project/', args,context)
else:
args = {}
args.update(csrf(request))
args['users'] = user
args['projects'] = projects
return render_to_response('Bapp/manage_projects.html', args,context)
def view_project(request, project_id):
context = RequestContext(request)
user = User.objects.get(project_sites__id=project_id)
site = ProjectSite.objects.get(id=project_id)
args = {}
args.update(csrf(request))
args['Users'] = user
args['Project'] = site
return render_to_response('Bapp/view_project.html', args,context)
答案 0 :(得分:4)
这可以解决您的问题:
from django.shortcuts import redirect
from django.core.urlresolvers import reverse
if request.method == 'POST':
project_id = request.POST.get('project_id')
return redirect(reverse('view_project', args=(project_id,)))
BTW另一个建议,而不是像我使用的那样使用Bapp/view_project.html
使用named URL patterns对网址路径进行硬编码。