django-ajax-选择示例用法

Question

任何人都可以发布django-ajax-choices最简单的例子吗？只有一个带有字段和表单的模型可用于根据该字段选择此模型的实例(不是在管理员中,而是在用户表单中).

我尝试阅读项目的文档,但发现它很难理解...另外,我无法使示例应用程序工作(coudl没有找出什么是标签???):(

另外,如果您认为有一个比django-ajax选择最简单的解决方案,请告诉我.

谢谢!

Answer 1

这是一个简单的例子(来自github上的示例但未经过测试):

models.py:

class Person(models.Model):
    name = models.CharField(blank=True, max_length=100)
    email = models.EmailField()

    def __unicode__(self):
        return self.name


class Group(models.Model):
    name = models.CharField(max_length=200,unique=True)
    members = models.ManyToManyField(Person,blank=True,help_text="Enter text to search for and add each member of the group.")

    def __unicode__(self):
        return self.name

forms.py:

class GroupForm(ModelForm):

    class Meta:
        model = Group

    members  = make_ajax_field(Release,'members','person')

lookups.py:

class PersonLookup(LookupChannel):

    model = Person

    def get_query(self,q,request):
        return Person.objects.filter(name__icontains=q).order_by('name')

    def get_result(self,obj):
        return obj.name

    def format_match(self,obj):
        return self.format_item_display(obj)

    def format_item_display(self,obj):
        return u"%s" % escape(obj.name)

settings.py:

AJAX_LOOKUP_CHANNELS = {
     'person' : ('example.lookups', 'PersonLookup'),
}

views.py:

class Create(generic.CreateView):
    template_name = "create.html"
    form_class = GroupForm
    success_url = 'create'

create = Create.as_view()

urls.py:

urlpatterns = patterns('',
    url(r'^create',  view='views.create',name='create'),
    url(r'^ajax_lookup/(?P<channel>[-\w]+)$', 'ajax_select.views.ajax_lookup', name = 'ajax_lookup'),
)