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How to show/hide menu item based on system parameter?


I've this code to create menu item, and it working fine, but i need to hide it depends on system parameter.


Does anyone know how to do that? Thanks.


<!-- this is for test, remove this after testing -->
<record id="check_qty_and_move_from_internal_customer_to_customer" model="ir.actions.server" style="color:#e8bf6a;">>
<field name="name">Cron Run Directly 2</field>
<field name="model_id" ref="model_kiotviet_cron"/>
<field name="binding_model_id" ref="onnet_kiotviet.model_kiotviet_cron"/>
<field name="state">code</field>
<!-- function called -->
<field name="code">model.check_qty_and_move_from_internal_customer_to_customer()</field>
</record>

<!-- this is for test, remove this after testing -->
<menuitem id="check_qty_and_move_from_internal_customer_to_customer" name="Cron Run Directly 2" parent="kiotviet_menu_root"
action="check_qty_and_move_from_internal_customer_to_customer" sequence="6"/>

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Автор Лучший ответ

After followed Niyas Raphy answer, i've solved my problem, and here's the code:


from odoo import models, api, tools, _


class Menu(models.Model):
_inherit = 'ir.ui.menu'

@api.model
@tools.ormcache('frozenset(self.env.user.groups_id.ids)', 'debug')
def _visible_menu_ids(self, debug=False):
menus = super(Menu, self)._visible_menu_ids(debug)
# get system parameter
if not self.env['ir.config_parameter'].sudo().get_param('your_module.your_config_name'):
# get menu item id
menu_item_id = self.env.ref('your_module.menu_item_id').id
# remove it from menus
menus.discard(menu_item_id)
return menus

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Hi,

You can inherit the _visible_menu_ids function from the ir.ui.menu model and filter the menus based on the configuration in the system parameters.

An use case can be found in this module: https://apps.odoo.com/apps/modules/15.0/kg_hide_menu/


Thanks

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thanks you sir, i will try this and will update the result on this post.

Автор

i've followed your answer and found the answer for my self, thanks you sir.

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