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How can I call a wizard from another module/folder??

I know where are the view and the .py what can I do for call that wizard on another menu?

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Ignorer

Which one do you want to call? Please tell me the path and i'll give you an answer according your wizard.

Meilleure réponse

If you wanted to call a wizard from button you need to write a button like

<button name='%(module_name.window_action)d' type='action' string='Button'/>

and if you have wrote a method in py file then you need to return action like

def your_method(self, cr, uid, ids, context=None):
    .....
    your code
    .....
    return {
       'name': 'Wizard name'
       'type': 'ir.actions.act_window',
       'res_model': 'wizard_model',
       'view_mode': 'form',
       'view_type': 'form',
       'target': 'new',
       'context': context,
    }

In return "target: new" will popup a wizard. This will help you to call a wizard from another module.

Avatar
Ignorer
Auteur

I did that but I have an error

raise NotImplementedError('Many2Many columns should not be used as record name (_rec_name)') NotImplementedError: Many2Many columns should not be used as record name (_rec_name)

Auteur Meilleure réponse

I did that but I have an error

raise NotImplementedError('Many2Many columns should not be used as record name (_rec_name)') NotImplementedError: Many2Many columns should not be used as record name (_rec_name)
What's Happen?

Avatar
Ignorer

I suppose you have defined many2many field and in that you are giving columns as name field of model, change it by suffix "id" it will work. for eg. model_id.

I suppose you have defined many2many field and in that you are giving columns as name field of model, change it by suffix "id" it will work. for eg. model_id.

Auteur

thanks it works now

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