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rosey

I have two fields, both are many2one

'dest_location' : fields.many2one('location.location',"New Location"),
'new_project' : fields.many2one('project.project','New Project'),

 <field name="dest_location" on_change="onchange_project_id(dest_location)"/>
 <field name="new_project" on_change="onchange_location_id(new_project)" />

My doubt :

Error in

def onchange_project_id(self, cr, uid, id, dest_location, context=None):
       if dest_location:
       project = self.pool.get('location.location').browse(cr, uid, dest_location, context=context).location_ids
       return {'value': {'new_project': project.name}}

How to return one2many field in a onchange function?

class location_location(osv.osv):

      _name = 'location.location'
      _description = 'Site Information'
      _columns = {
                  'name': fields.char('Location'),
                  'location_ids': fields.one2many('project.location', 'project_id', 'Work done'),
        
    }
location_location()

class project_location(osv.osv):
      _name = "project.location"
      _description = "Project Location"
      _columns = {
        
        'project_id': fields.many2one('project.project', 'Project'),
      
    }

How can i import all projects in the location table when onchange works?

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Anil Kesariya
Best Answer

Hello rosey,

You need to make list of tuples for your one2many field and return it in your result dictionary with one2many field

Here you go!

return {'value': {''location_ids'': [(0, 0,  { fields }), (0, 0,  { fields })]}

Hope this will help you.

Regards,

Anil.

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rosey
Author

Thanks Anil. I will try.

rosey
Author

its not working

Anil Kesariya

Use this format (0, 0, { fields }) list tuples. each dictionary will contain your line data.

rosey
Author

def onchange_project_id(self, cr, uid, id, dest_location, context=None): location = self.pool.get('location.location').browse(cr, uid, dest_location, context=context) print "location",location if dest_location: return {'value': {'location_ids': [{location.project_id.name}]}} return {'value': {}}

rosey
Author

is this correct??

Anil Kesariya

Nop. let say you have three fields inside project.location model like name,date now your return value for number location_ids will be like this. return {'value': {'location_ids': [(0,0,{'name':"project_first", 'date':"project_date"}), (0,0, {'name':'project_second', 'date':"date_second"})] }} so on..

Anil Kesariya

Nop. let say you have two fields inside project.location model like name,project_id now your return value for number location_ids will be like this. return {'value': {'location_ids': [(0,0,{'name':"project_first", 'project_id':id of your project.}), (0,0, {'name':'project_second', 'project_id':id of your project })] }} so on..

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Gelson
Best Answer 
terms_obj = self.env['x.x'].search([('terceiro_id', '=', self.terceiro_id_doc_conta_corente.id),('pago1', '=', False)])
list_of_dict = []para linha em terms_obj: 
     data = self ._comput_line (linha) 
     data.update ({ ' name1 ' : line.name1 , 'pago1' : line.pago1 , ' date1 ' : line.date , 'acobrado1' : line.acobrado1 , 
                     'totalr1' : line.totalr , 'saldo1' : line.saldo}) 
list_of_dict.append ((0 , 0 , data)) return { 'valor' : { "doc_cont_corent" : list_of_dict}}                  

this will help you

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Ben Bernard
Best Answer

return {'value': {'new_project': project.name}}

project is plural, but new_project is singular, so which project should return? If you want to return the first project, then

return {'value': {'new_project': project[0].id}}

Hmm, I think I should edit also the other answer from related question -___-

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rosey
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i want to return all projects, which have location the same. If location 1 have 3 projects, i want to select all of the three only in the list.

Ben Bernard

You can't do it, because new_project is many2one.