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insertsql
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Sebastian782

Hello,

I have 2 models A and B.

When i update field_1 on A model i execute a function that should add a new record into B model.

How can i do this?

I read that we can execute SQL request (example belaw) but maybe there is an other way more clean than do an INSERT to add new record?

query = """ SELECT event_id, state, sum(nb_register) 
            FROM event_registration 
            WHERE event_id IN %s AND state IN ('draft', 'open', 'done') 
            GROUP BY event_id, state """ 
self._cr.execute(query, (tuple(self.ids),))

Thanks,
Sebastian

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Akhil P Sivan
最佳答案

You can override the write function of model_A for that, so that when you are saving the record, it create a new record on model_B.

For eg, try the following, if you want to use old api in your v8 module:

from openerp.osv import fields, osv
class class_A(osv.osv):
    _name = 'model.A'
    _columns = {
        'field_A': fields.many2one('res.partner', string="Supplier", required=True),
     
    }
    
    def write(self, cr, uid, ids, vals, context=None):
        
        model_B_obj = self.pool.get('model_B')
        model_B_obj.create(cr, uid, ids, {'product_id':ids, 'date':vals['status_date'], 'status':vals['status']}, context)
        
        return super(class_A, self).write(cr, uid, ids, vals, context)

Using new api, you may try like this:

class class_A(models.Model):
    _name = 'model.A'
    
    field_A = fields.Many2one('res.partner', string="Supplier", required=True)
     
    
     
    @api.multi
    def write(self, vals):
        
        model_B_obj = self.env['model_B'].create({'product_id':self.id, 'date':self.status_date, 'status':self.status})
        
        return super(class_A, self).write(vals)

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Sebastian782
作者

Hi, thanks. I don't know why, i am not able to make it work. When i change my select field, it seems that it create a record, with the default values of model.B. First, I want to create record in model.B when i save model.A record, so i think the onchange is not good? It seem that it create a record when Second, i want to create the model.B with my values status_date and status of model.A Here below what i wrote: ------------------------------- @api.onchange('status') def on_change_status(self): self.env['order.history'].create({'product_id':self.id, 'date':self.status_date, 'status':self.status}) What i am doing wrong? Thanks.

Akhil P Sivan

Hi Sebastian, I have updated the answer. You don't need to use onchange for that, just need to override the write function of model_A. Try like above.

Sebastian782
作者

Hi Akhil, thanks for your help. I have this error : TypeError: write() takes at least 2 arguments (2 given)

Sebastian782
作者

I use the new API

Akhil P Sivan

try without context, as updated above

Sebastian782
作者

Perfect now Akhil, thanks :)

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james_p
最佳答案

You can achieve it without firing SQL request. Just add the below method in model A. You can read more here: https://www.odoo.com/documentation/8.0/reference/orm.html Hope that helps.

@api.onchange('field_1')
def create_new_record(self):
    self.env['model_b'].create({'field_in_model_b': value})
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Sebastian782
作者 最佳答案

Hi, thanks.

I don't know why, i am not able to make it work.

When i change my select field, it seems that it create a record, with the default values of model.B.

First, I want to create record in model.B when i save model.A record, so i think the onchange is not good? It seem that it create a record when

Second, i want to create the model.B with my values status_date and status of model.A

Here below what i wrote:

-------------------------------

@api.onchange('status')

def on_change_status(self):

self.env['order.history'].create({'product_id':self.id, 'date':self.status_date, 'status':self.status})

What i am doing wrong?

Thanks.

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