Odoo Server Action python code open new modal popup - popup

Question

I created Automated action for my partner table on creations add a server action than I want to the server action open a new form on modal popup but I don't write this python code.
Anybodys help me please...

Answer 1

You can add a button, as Yogesh said, but you'll want to make it type object. This will link to a function in the tree's model class, and this function can return a new view. For example, in the xml:
Then in the class definition for this model:
#api.multi
def button_details(self):
view = {
'name': _('Details'),
'view_type': 'form',
'view_mode': 'form',
'res_model': 'my.custom.model',
'view_id': False,
'type': 'ir.actions.act_window',
'target': 'new',
'readonly': True,
'res_id': self.id,
}
return view
Make sure to replace 'my.custom.model' with the actual model you want to edit. Also, you'll need to have already definied the ir.actions.act_window and ir.ui.view records for that model, or Odoo won't know how to bring up the details. That top line there is from the new v8 api, but if you're using v7, remove it and just add the standard args to the button_details function call (cr, uid, ids, context). There are lots of examples of button functions in the code, I'm sure you can find one. The key to the button bringing up a new modal window is the line :
'target': 'new',
This will make the details edit be its own little popup window. Alternativly, you can set target to 'current' to just redirect to that page normally, but I'm going to guess in this instance you'd rather a little popup and then return to the main view.
The "'res_id': self.id," line is very important, and I think it only works with the v8 API. If you're using v7, instead of "self.id", you'll probably want the "ids" arg passed in from the function arguments. I hope that makes sense.

Answer 2

#api.multi
def button_approved(self):
self.ensure_one()
return { 'name': 'My Name',
'type': 'ir.actions.act_window',
'views':[[False,'form']],
'res_model': 'my.custom.model',
'target': 'new',
'context': {'default_my_type':'state','default_model_id':self.id} }
but this code for button. I want to automatic actions for when created my.partner.model Your code doesn't work for server actions python code

Answer 3

As far as I understood you need to create a wizard using only xml files so that you can upload on saas, so here after creating automated action then you need to create a view and action and bind that action with your server action.
View:
<record model="ir.ui.view" id="view_name_wizard">
<field name="name">name.wizard.form</field>
<field name="model">model_name</field>
<field name="priority">16</field>
<field name="arch" type="xml">
your fields # remember to add new model and fields from xml
</field>
</record>
Action:
<record id="action_name_wizard" model="ir.actions.act_window">
<field name="name">Title</field>
<field name="type">ir.actions.act_window</field>
<field name="target">new</field>
<field name="res_model">model_name</field>
<field name="view_type">form</field>
<field name="view_mode">form</field>
<field name="view_id" ref="view_name_wizard"/>
</record>
Now say if you want to popup this wizard on automated action you need to add this action on server action with this. action_name_wizard .
Tell me if you need anything more.