How remove user from multiple groups at once Odoo 12

Odoo is the world's easiest all-in-one management software.
It includes hundreds of business apps:

CRM
e-Commerce
Kirjanpito
Varastointi
PoS
Project
MRP

All apps

Kaikki kirjoitukset Ihmiset Merkit

Tunnisteet (Näytä kaikki)

odoo accounting v14 v15 pos

Tietoa tästä foorumista

Apua

How remove user from multiple groups at once Odoo 12

user groups odoo odoo12

2 Vastaukset

6391 Näkymät

Pawan Kumar Sharma

Hello All,

I want to remove the user from multiple groups at once when changing from an 'individual' to 'company' on write method of a contact form.

@api.multi
def write(self, values):
    user = self.env['res.users'].search([('partner_id', '=', self.id)])
    if 'company_type' in values:
        if values["company_type"] == "company":
            # want to add in below group
            """[(6, 0, [self.env.ref('base.group_user').id,
                                           self.env.ref('sales_team.group_sale_salesman_all_leads').id,
                                           self.env.ref('survey.group_survey_manager').id
                                           ])]"""

            # want to remove from below group
            """[(6, 0, [self.env.ref('base.group_portal').id,
                                           self.env.ref('survey.group_survey_user').id,
                                           ])]"""
    res = super(Contact, self).write(values)
    return res

Thanks in advance

Malay Khamar (Serpent Consulting Services Pvt. Ltd.)

Paras vastaus

Hello,

You can try this,

@api.multi
def write(self, values):
    user = self.env['res.users'].search([('partner_id', '=', self.id)])
    if 'company_type' in values:
        if values["company_type"] == "company":
            # want to add in below group
            user.write({'groups_id': [(4, self.env.ref('base.group_user').id),
                                                    (4, self.env.ref('sales_team.group_sale_salesman_all_leads').id),
                                                    (4, self.env.ref('survey.group_survey_manager').id)
                                                       ]})
            # want to remove from below group
           user.write({'groups_id': [(3, self.env.ref('base.group_portal').id),
                                                    (3, self.env.ref('survey.group_survey_user').id),
                                                        ]})
    res = super(Contact, self).write(values)
    return res

You can also use like this:

    user.write({'groups_id': [(4, self.env.ref('base.group_user').id),
                                            (4, self.env.ref('sales_team.group_sale_salesman_all_leads').id),
                                            (4, self.env.ref('survey.group_survey_manager').id),
                                            (3, self.env.ref('base.group_portal').id),
                                            (3, self.env.ref('survey.group_survey_user').id),
                                           ]})

Hope this will work for you,

Thanks

Pawan Kumar Sharma

Tekijä

Thanks Malay, it's working fine

Avinash Nk

Hello Pawan,

To remove a user from a group, you can do it like this

 your_group = self.env.ref('module.odoo_group', False)
 your_group.write({'users': [(3, user_id)]})


For Adding,

 your_group = self.env.ref('module.odoo_group', False) 
 your_group.write({'users': [(4, user_id)]})

You can use loop and your logic to execute all at once.


Check this for more understanding, how to write to a One2many field.

(0, 0, { values })

link to a new record that needs to be created with the

given values dictionary

(1, ID, { values })

update the linked record with id = ID (write values on it)

(2, ID)

remove and delete the linked record with id = ID

(calls unlink on ID, that will delete the object completely,

and the link to it as well)

(3, ID)

cut the link to the linked record with id = ID

(delete the relationship between the two objects but does

not delete the target object itself)

(4, ID)

link to existing record with id = ID (adds a relationship)

(5)

unlink all (like using (3,ID) for all linked records)

(6, 0, [IDs])

replace the list of linked IDs (like using (5) then (4,ID)

for each ID in the list of IDs)




Thanks & Regards
Avinash N K

Pawan Kumar Sharma

Tekijä

Thanks avinash, it's working

Avinash Nk

My pleasure

Nautitko keskustelusta? Älä vain lue, vaan osallistu!

Luo tili jo tänään nauttiaksesi yksinoikeusominaisuuksista ja osallistuaksesi mahtavaan yhteisöömme!

Rekisteröidy

Aiheeseen liittyviä artikkeleita	Vastaukset	Näkymät
Odoo Users Creation user odoo	1 marrask. 23	921
How to track One2many field in Odoo12? Ratkaistu odoo odoo12	3 marrask. 23	16441
Error: Couldn't load module web Ratkaistu odoo odoo12	3 marrask. 24	23507
How I get the current ID of a record ? Ratkaistu odoo odoo12	1 huhtik. 23	5430
Binary field as attachment in mail in Odoo 13 Ratkaistu odoo odoo12	2 jouluk. 22	6113

Tämä kysymys on merkitty

Nautitko keskustelusta? Älä vain lue, vaan osallistu!