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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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破棄
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Anil Kesariya
最善の回答

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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破棄
rosey
著作者

Thanks Anil. I will try.

rosey
著作者

its not working

Anil Kesariya

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

rosey
著作者

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
著作者

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
最善の回答 
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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破棄
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Ben Bernard
最善の回答

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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破棄
rosey
著作者

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.