ExpertExpertThat only works for the immediate group. I need subgroups included as well. Since websites don't contain the FullPath field like resources do, I don't think there is any way to do this.
LegendYeah, you'd just have to loop through all groups:
website_groups = lm.get_website_group_list(size=1000).items
for group in website_groups:
websites = lm.get_website_list(size=1000, filter=f"groupId:{group.id}").items
print(f"{group.full_path}: {','.join([x.name for x in websites])}")
LegendOne liner if you play code golf:
for group in lm.get_website_group_list(size=1000).items: print(f"{group.full_path}: {','.join([x.name for x in lm.get_website_list(size=1000, filter=f'groupId:{group.id}').items])}")
LegendIf you want the list per root group:
all_websites = {group.full_path: ','.join([x.name for x in lm.get_website_list(size=1000, filter=f'groupId:{group.id}').items]) for group in lm.get_website_group_list(size=1000).items}
by_root_group = {}
for full_path, websites in all_websites.items():
rootpath = full_path.split("/")[0]
if rootpath in by_root_group.keys():
by_root_group[rootpath] += websites
else:
by_root_group[rootpath] = websites
for k,v in by_root_group.items(): print(f"{k}: {v}")
