带有列表的Django URL模式(Django URL pattern with a list)

带有列表的Django URL模式(Django URL pattern with a list)

我有一个包含类别名称的列表，例如cats = ["tv", "movie", "theater"] 。 我想写一个url模式来只捕获包含列表中某个项的URL，例如：

url(r'^site/CATEGORY_NAME/$', 'mainsite.views.home'),

这样CATEGORY_NAME只能列出cat中的一个项目。 我怎样才能做到这一点？

谢谢，梅尔

I have a list with category names , e.g. cats = ["tv", "movie", "theater"]. I would like to write a url pattern to catch only URLs which contain one of the items in the list, such as:

url(r'^site/CATEGORY_NAME/$', 'mainsite.views.home'),

so that CATEGORY_NAME can only one one of the items in the list cats. How can I do that?

Thanks, Meir

最满意答案

您可以使用python的字符串join方法从列表中构建正则表达式的一部分，然后在URL模式中使用它。

例如：

cats = ["tv", "movie", "theater"] cats_re = '(?:' + '|'.join(cats) + ')' # ...then... url(r'^site/' + cats_re + '/$', 'mainsite.views.home'),

在这种情况下，整个正则表达式看起来像：

url(r'^site/(?:tv|movie|theater)/$', 'mainsite.views.home'),

You can build part of a regular expression from the list by using python's string join method, and then use that in the URL pattern.

For example:

cats = ["tv", "movie", "theater"] cats_re = '(?:' + '|'.join(cats) + ')' # ...then... url(r'^site/' + cats_re + '/$', 'mainsite.views.home'),

In this case, the whole regular expression would look like:

url(r'^site/(?:tv|movie|theater)/$', 'mainsite.views.home'),