django和python的新手,请以非常简单的方式向我提出建议.
基本上我有一个图像托管网站,我希望能够点击任何图像,让它打开一个新页面显示图像.我该怎么做呢？我认为我必须得到图像网址,但我不知道如何.这也必须是自动的.
楷模:
class UploadImages(models.Model):
category_choice=(('Full','Full Beard'),
('tashe', 'Moustache'),
('goat','Goatee'))
image = models.ImageField(upload_to='rate', blank=True)
name = models.CharField(max_length=128, default=0)
user = models.ForeignKey(User)
date = models.DateTimeField(default=now())
#category = models.ForeignKey(Categories)
category = models.CharField(max_length=30, choices=category_choice)
rating = models.IntegerField(default=0)
def __unicode__(self):
return self.name
浏览次数:
def index(request):
if request.method == 'GET':
#images_list = UploadImages.objects.order_by('-date')
images_list = UploadImages.objects.filter(category='tashe')
category_list = Category.objects.order_by('-likes')[:5]
page_list = Page.objects.order_by('-views')[:5]
context_dict = {'images':images_list}
response = render(request,'RateMyBeard/index.html', context_dict)
return response
else:
rating = request.POST['submit']
#images_list = UploadImages.objects.order_by('-date')
images_list = UploadImages.objects.filter(category='tashe')
print rating
#print images_list …