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actiondevelopementvisibilityserver_actions17.0
3 Replies
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Master

Hi,

How can I show or hide a server action button based on a field value in my model?

For example, if the state field is 'closed', I want the "Open" button to be visible.

And if the state is 'open', I want the "Close" button to be displayed instead.

Is there a way to conditionally show or hide these server actions depending on the record's state?


<record id="mymodel_open_action_server" model="ir.actions.server">

<field name="name">Open mymodel</field>

<field name="model_id" ref="model_mymodel"/>

<field name="binding_model_id" ref="model_mymodel"/>

<field name="binding_view_types">form</field>

<field name="state">code</field>

<field name="code">

action = records.open_action()

</field>

Avatar
Buang
Avatar
Cybrosys Techno Solutions Pvt.Ltd
Jawaban Terbai

Hi,

Please refer to the solved forum:

1. https://www.odoo.com/id_ID/forum/help-1/how-to-hide-action-button-based-on-condition-246523

2. https://www.odoo.com/id_ID/forum/help-1/hide-a-menu-inside-action-button-based-on-condition-168198


Hope it helps.

Avatar
Buang
Avatar
slckatav
Jawaban Terbai

Hey, could you share your open_action() function as well? Since you’d prefer not to use a button seeing that code will help me suggest the right approach. In general you can add an attrs to your XML to toggle visibility. For example:

attrs="{'invisible': [('state', 'in', ['sale'])]}"

This will hide the element when state is not in the “sale” stage. Of course, you can adjust the domain or use different attribute just a quick example to get you started.

Avatar
Buang
Master
Penulis

Hi, you can see my code below for more details:

<!-- Server Action -->
<record id="mymodel_closed_action_server" model="ir.actions.server">
<field name="name">Close</field>
<field name="model_id" ref="model_mymodel"/>
<field name="binding_model_id" ref="model_mymodel"/>
<field name="state">code</field>
<field name="code">action = records._action_close()</field>
</record>

def _action_close(self):
for rec in self:
if rec.state != "closed":
print("My logic goes here!")
rec.state = "closed"

slckatav

I once hid some action buttons by finding their generated names. Odoo assigns its own names to action buttons like button name="356". In your case, after you add your action button find the name that Odoo generated and then add XML like this:

<xpath expr="//button[@name='466']" position="attributes">
<attribute name="invisible">True</attribute>
</xpath>

I hope this helps!

Avatar
D Enterprise
Jawaban Terbai

Hiii,

To show/hide buttons based on the state field in Odoo 17 Community Edition , you must use button elements inside the form view , not ir.actions.server bindings, because ir.actions.server buttons do not support dynamic visibility using attrs .

Here's the correct and clean implementation :

Model: mymodel.py

Define state , action_open , and action_close :

from odoo import models, fields

class  MyModel (models.Model):
    _name = 'mymodel' 
    _description = 'My Model'

    state = fields.Selection([
        ( 'open' , 'Open' ),
        ( 'closed' , 'Closed' )
    ], default= 'open' , string= 'State' )

    def  action_open ( self ):
         for record in self:
            record.state = 'open'

    def  action_close ( self ):
         for record in self:
            record.state = 'closed'
View: mymodel_form_view.xml

Inherit or define your form view and add two buttons, conditionally shown using attrs .

<record id ="view_form_mymodel"  model="ir.ui.view">
     <field  name = "name" >mymodel.form </field>
     <field  name = "model" >mymodel </field>
     <field  name = "arch"  type = "xml">
         <form  string = "My Model">
             <header>
                 <button name="action_open"
	​	​	​	​	​        type="object"
	​	​	​		​​        string="Open"
        	​	​	​	​	​invisible="state != 'closed'"/>
                <button name="action_close"
	 ​	​	​	​	​        type="object"
	​	​	​	​	​        string="Close"
	​	​	​	​	​        invisible="state != 'open'"/> 
             </header>

            <sheet>
                 <group>
                     <field  name = "state"/>
                     <!-- other fields --> 
                </group>
             </sheet>
         </form >
     </field >
 </record>

i hope it is use full

Avatar
Buang
Master
Penulis

This works with a normal <button> tag. However, in my case, I need to show or hide my custom choices Close/Open in the default server action, similar to the behavior of the Archive/Unarchive button in

D Enterprise

Use the binding_filter_domain field (Best Practice)
From Odoo 15+, ir.actions.server supports a field called binding_filter_domain, which defines when a server action appears in the action dropdown.

Example: Show “Open” action only if state == 'closed'
<record id="mymodel_open_action_server" model="ir.actions.server">
<field name="name">Open</field>
<field name="model_id" ref="model_mymodel"/>
<field name="binding_model_id" ref="model_mymodel"/>
<field name="binding_view_types">form</field>
<field name="binding_filter_domain">[('state', '=', 'closed')]</field>
<field name="state">code</field>
<field name="code">
action = records.action_open()
</field>
</record>
Example: Show “Close” action only if state == 'open'
<record id="mymodel_close_action_server" model="ir.actions.server">
<field name="name">Close</field>
<field name="model_id" ref="model_mymodel"/>
<field name="binding_model_id" ref="model_mymodel"/>
<field name="binding_view_types">form</field>
<field name="binding_filter_domain">[('state', '=', 'open')]</field>
<field name="state">code</field>
<field name="code">
action = records.action_close()
</field>
</record>
please try this
i hoe it is use full

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