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lj

I am looking for the technique how to add specific functionality/action to a tree view.


in a tree view I have added a button

<tree>
    <button name="fn1" type="object" string="F1" />


then I have added a method for this button in my model


def fn1(self):

All fine to this point


Problem 1 (from view to a method)

I assume that when I click on a button on the record I would somehow get the record id in which button is clicked on. But...


Normally models method def would have REC as parameter and would consist of a marked record set, but in this case no record set has been passed to a method. How do I know on which record button I have clicked on? Is there some modifier in the view definition to instruct button to send record id to an action method?


Problem 2 (from a method to a resulting page)

There is technique to navigate to a specific form using this kind of button, by making method return data in this fashion:

return {
    'view_type': 'form',
    'view_mode': 'form',
    'res_model': 'my.cloud.files',
    'views': [(False, 'form')],
    'type': 'ir.actions.act_window',
    'target': 'new',
    'context': {},
}

this renders a popup form (in described case it is not even a record from which click is originated but it looks like 'create' form), apart from this:

I do not want to use any of existing ODOO forms, what I want to navigate to a dynamic url I am going to build within this model method.

In more detail what I want is to construct URL from the record I have clicked on, and navigate browser to that URL i have built. To build URL I will be using using a field from my model. For example I have field in the model which is called ID and it has value 123 so I would build a URL like http://site/path/123. This is trivial to make in python but later instruction to ODOO to go to that 'dynamic' url is something I cannot find how to do. 

how should 'return' look like to go to my url ?


Kindly advise.

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lj
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Đức Hậu Trần

Thank you for your answer. Need to try these hints.

One small clarification, please. Following code looks to me to be a part of a Model definition. Is this correct understanding?

fb_link = fields.Html ( string= "FB link", compute="get_fb_link" )

(have not enough karma to comment on your post)

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Đức Hậu Trần
Best Answer

Hi,

1. you can use html dynamic field to define your url and show it in your form.

for example:


#Python code:

def get_fb_link( self):

    for item in self:

        item.fb_link = "http://site/path/" + item.id

fb_link = fields.Html ( string= "FB link", compute="get_fb_link" )


# in xml file, you show fb link in your form.


2. You can use "ir.actions.act_url" model to return your URL.

#https://www.odoo.com/documentation/13.0/reference/actions.html#url-actions-ir-actions-act-url

def fn1(self):

return {
    "type": "ir.actions.act_url",
    "url": "http://site/path/" + self.id,
    "target": "self",
}

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