如何在odoo10中點擊打開新窗體的按鈕來解決錯誤？

Question

def create_file(self): 

    opportunity_id = self.convert_to_file() 
    return self.env['trademark.process'].view_file(opportunity_id) 

我用轉換文件功能，目前的模型的一些值傳遞給trademark.process

def convert_to_file(self, partner_id=False): 

    tm_process_obj = self.env['trademark.process'] 
    tm_search_record = self.env['trademark.search'].browse(self.id) 
    for rec in tm_search_record: 
     opportunity_id = tm_process_obj.create({ 
         'search_name_char': rec.search_name or False, 
         'classification_no_id':rec.classification_no_id.id or False, 
         'partner_id': rec.partner_id.id or False, 
         'user_id': rec.user_id.id or False, 
         'search_date': rec.search_date or False, 
         'search_seq_no': rec.seq_no or False, 
        }) 
     vals = { 
      'file_no': opportunity_id.seq_no, 
     } 
     self.write(vals) 
    return opportunity_id 

然後我終於回到機會ID，並傳遞給視圖文件的功能。

def view_file(self, opportunity_id): 
    view_id=self.env.ref('trademark_services.trademark_process_form_view').id 
    return { 
     'name': _('File Details'), 
     'type': 'ir.actions.act_window', 
     'view_type': 'form', 
     'view_mode': 'form', 
     'res_model': 'trademark.process', 
     'view_id': view_id, 
     'res_id': opportunity_id, 
     'target':'current' 
    } 

但是當我點擊按鈕時發生錯誤。

Traceback (most recent call last): 
File "/home/ubuntu/workspace/amzsys_erp/odoo/http.py", line 638, in 
_handle_exception 
return super(JsonRequest, self)._handle_exception(exception) 
File "/home/ubuntu/workspace/amzsys_erp/odoo/http.py", line 689, in 
    dispatch 
    return self._json_response(result) 
File "/home/ubuntu/workspace/amzsys_erp/odoo/http.py", line 627, in 
_json_response 
    body = json.dumps(response) 
File "/usr/lib/python2.7/json/__init__.py", line 243, in dumps 
    return _default_encoder.encode(obj) 
File "/usr/lib/python2.7/json/encoder.py", line 207, in encode 
chunks = self.iterencode(o, _one_shot=True) 
File "/usr/lib/python2.7/json/encoder.py", line 270, in iterencode 
    return _iterencode(o, 0) 
File "/usr/lib/python2.7/json/encoder.py", line 184, in default 
    raise TypeError(repr(o) + " is not JSON serializable") 
TypeError: trademark.process(131,) is not JSON serializable 

如何解決此問題。單擊按鈕時如何打開新窗體。

我想將某些值傳遞給該窗體。

我的代碼有什麼錯誤？

注意：使用odoo10

Answer 1

問題在您傳遞res_id的方法中。

使用opportunity_id.id。

def view_file(self, opportunity_id): 
    view_id=self.env.ref('trademark_services.trademark_process_form_view').id 
    return { 
     'name': _('File Details'), 
     'type': 'ir.actions.act_window', 
     'view_type': 'form', 
     'view_mode': 'form', 
     'res_model': 'trademark.process', 
     'view_id': view_id, 
     'res_id': opportunity_id.id, 
     'target':'current' 
    }