I want to use a database query to generate my URL configuration. Something like:
我想使用数据库查询来生成我的URL配置。就像是:
states = State.objects.all().values_list('pk', flat=True)
And then a regex like:
然后像正则表达式:
(r'^state/(?P<state>' + '|'.join(states) + ')/$'
The idea is that I can have URLs like:
我的想法是,我可以拥有以下网址:
/state/ca/
/state/az/
Etc.
The problem is that when I do syncdb, the query above fails, throwing a DatabaseError.
问题是,当我执行syncdb时,上面的查询失败,抛出DatabaseError。
Is there a good way to do this? I've tried the obvious change, which would be:
有没有办法做到这一点?我尝试过明显的改变,那就是:
try:
states = State.objects.all().values_list('pk', flat=True)
except DatabaseError:
# First sync, use dummy data
states = []
But this doesn't work because the exception is thrown at the regex, not at the query definition.
但是这不起作用,因为在正则表达式中抛出异常,而不是在查询定义中抛出异常。
Ideas?
1 个解决方案
#1
Why do you need to constrain this in the URL pattern itself? Much better to accept all two-letter codes, and check in the view.
为什么需要在URL模式本身中约束它?接受所有双字母代码要好得多,并在视图中查看。
(r'^state/(?P<state_code>\w{2})/$'
def view_state(request, state_code):
state = get_object_or_404(State, pk=state_code)
#1
Why do you need to constrain this in the URL pattern itself? Much better to accept all two-letter codes, and check in the view.
为什么需要在URL模式本身中约束它?接受所有双字母代码要好得多,并在视图中查看。
(r'^state/(?P<state_code>\w{2})/$'
def view_state(request, state_code):
state = get_object_or_404(State, pk=state_code)