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How to send url of a record by email ?

By
Anabela Damas
on 3/5/15, 9:53 AM 969 views

Hi, 

I've a object that has a user_id ('user_id': fields.many2one('res.users', u'Responsável', required=True,),), and this user recive an email when someone changes the state of the record. 

I've this function to give the url: 

      def get_signup_url(self, cr, uid, ids, context=None):
        document = self.browse(cr, uid, ids, context=context)
        res= False
        base_url = self.pool.get('ir.config_parameter').get_param(cr, uid, 'web.base.url')
     
        query = {'db': cr.dbname}
        fragment = {'action': 'vi_resource_reservation.action_resource_reservation',
                    'type': "signup",
                    'view_type' : 'form',
                    'model' : 'resource.reservation',
                    'id': document.id,
                    'login': document.user_id.login,
                    }
       
        res = urljoin(base_url, "?%s#%s" % (urlencode(query), urlencode(fragment)))
        print res
        return res

It return the following link, with the id 11 :

1) http://192.168.2.141:8086db=testes_0503#view_type=form&id=11&action=vi_resource_reservation.action_resource_reservation&login=user2&type=signup&model=resource.reservation

But when I write the link 1) in the browser it redirects to the first id in the database of the model (id 2 ):

2) http://192.168.2.141:8086/?db=testes_0503#id=2&view_type=form&model=resource.reservation&action=606

Do you have any ideia why? What I'm doing wrong?

thanks 

 

0
Mariusz Mizgier
On 3/5/15, 10:36 AM

1) Your URL is wrong, as it should be like:

http://192.168.2.141:8086/db=testes_0503#view_type=form&id=11

&action=vi_resource_reservation.action_resource_reservation&login=user2&type=signup&model=resource.reservation

2) If point 1 doesn't work, try passing the link without action parameter.

the 2) works. Thanks =)

Versão Integral, Anabela Damas
on 3/5/15, 1:07 PM

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Asked: 3/5/15, 9:53 AM
Seen: 969 times
Last updated: 3/16/15, 8:10 AM