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shaneebkhan

how to add a button in tree view to get the form view of the desired record(as the list view is in editable mode,it wont directly go to the form view)?.how to set the action of the button to open the edit mode of form view?

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Axel Mendoza

not really useful since you can open the form by clicking directly on the record line, why do you need a button in the record to do specifically the same? or in the top buttons of the tree where you can select multiple record lines?

shaneebkhan
Autor

i actually changed the list view as editable ,so it wont go to the form view directly when click.so i would like to add a edit button next to every record,so when i click on edit,it should go to the edit part of form view.

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Ray Carnes
Best Answer

1. Add a function (something like this) to the model that opens the form

def open_to_form_view(self, cr, uid, ids, context=None):
 
    if not context:
        context = {}

    name = '<Name>'
    res_model = '<model>' 
    view_name = '<view>' 
    
    document_id = self.browse(cr, uid, ids[0]).id

    view = models.get_object_reference(cr, uid, name, view)
    view_id = view and view[1] or False


    return {
        'name': (name),
        'view_type': 'form',
        'view_mode': 'form',
        'view_id': [view_id],
        'res_model': res_model, 
        'type': 'ir.actions.act_window',
        'nodestroy': True,
        'target': 'current',
        'res_id': document_id,
    }

 

2. Add a button to your treeview like this:

<button name="open_to_form_view" type="object" icon="STOCK_INDEX"/>

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rodc
Best Answer

can u explain deeper about the

    name = '<Name>'
 res_model = '<model>'  
 view_name = '<view>'
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Annadurai

name as you give.

res_model like sale.order, product.product (table name)

view_name --> sale.view_order_form (sale order form view)

Annadurai

view_name xml id of sale order form view.

rodc

sorry about the very late reply, Thank you for explaining I understand it now