[FIX] website: perf - prefetch every website menus in one go

Without this commit, the website menus would not be prefetch all in once.
Thus, every call to `submenu` template (called recursively) would do SQL
queries. One when accessing website.menu_id, then for every access on menu's
child_ids. It would also make an extra request for groups since the menus have
an ir.rule for that.

with following structure, 13 SQL queries will be avoided, regardless if there is
a group or not on the menus:
- Top menu
  - Home
  - Contact us
    - Sub Contact us
      - Sub Sub Contact us

Related to #47257
task-2211013

X-original-commit: 7fb016f9966bb5e2289ba17eacafb8887626514e

1 files changed, 11 insertions, 2 deletions

diff --git a/addons/website/models/website.py b/addons/website/models/website.py
index c8dd03d8107..270a6199c95 100644
--- a/addons/website/models/website.py
+++ b/addons/website/models/website.py
@@ -133,9 +133,18 @@ class Website(models.Model):
             self.default_lang_id = language_ids[0]
 
     def _compute_menu(self):
-        Menu = self.env['website.menu']
         for website in self:
-            website.menu_id = Menu.search([('parent_id', '=', False), ('website_id', '=', website.id)], order='id', limit=1).id
+            menus = self.env['website.menu'].search([('website_id', '=', website.id)])
+
+            # use field parent_id (1 query) to determine field child_id (2 queries by level)"
+            for menu in menus:
+                menu._cache['child_id'] = ()
+            for menu in menus:
+                # don't add child menu if parent is forbidden
+                if menu.parent_id and menu.parent_id in menus:
+                    menu.parent_id._cache['child_id'] += (menu.id,)
+
+            website.menu_id = menus and menus.filtered(lambda m: not m.parent_id)[0].id or False
 
     @api.model
     def create(self, vals):