Issue 5511 - Create Product Help Home Template for help.eyeo.com

globals/get_page_slug.py

Index: globals/get_page_slug.py
===================================================================
new file mode 100644
--- /dev/null
+++ b/globals/get_page_slug.py
@@ -0,0 +1,20 @@
+# This file is part of help.eyeo.com.
+# Copyright [C] 2016 Eyeo GmbH
+#
+# help.eyeo.com is free software: you can redistribute it and/or modify
+# it under the terms of the GNU General Public License as published by
+# the Free Software Foundation, either version 3 of the License, or
+# [at your option] any later version.
+#
+# help.eyeo.com is distributed in the hope that it will be useful,
+# but WITHOUT ANY WARRANTY; without even the implied warranty of
+# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+# GNU General Public License for more details.
+#
+# You should have received a copy of the GNU General Public License
+# along with help.eyeo.com.  If not, see <http://www.gnu.org/licenses/>.
+
+def get_page_slug(url):

[ire, 2017/08/15 15:09:22]

Here I was trying to get just the unique slug part of a url. Wasn't able to find
a method that already exists to do this. Is there one?

[juliandoucette, 2017/08/24 11:13:59]

Yes it's the page's "page" property.

[ire, 2017/08/28 14:54:47]

That includes the whole path (not including the hostname), what I'm looking for
is just the part for the particular page, not parent pages as well

+
+  split_url = url.split('/')
+  return split_url[ len(split_url) - 1 ]