How to hide UI Action based on field change in form without using DOM?

elanrajad · ‎02-21-2017

Hi Team,

We have check box field on form if check box is checked need to hide button without saving the form and if check box is unchecked need to show the button on form without using DOM .Any one Can help me..

Thanks

Elanraj

nitish99 · ‎02-21-2017

Hi Elanraja,

1) Create a UI Policy --> Set Condition as field (check_box) is true

2) In the Script tab set Run scripts true

3) In the if condition write

$$('#action_name')[0].hide(); //This will hide the top button

$$('#action_name')[1].hide(); //This will hide the bottom button as well

4) In the else condition write

$$('#action_name')[0].show(); //This will show the top button

$$('#action_name')[1].show(); //This will show the bottom button as well

For reference,

check this How to hide/show an UI action on field changes

View solution in original post

Prati · ‎01-20-2021

i am trying to achieve to hide ui action with help of above ui policy but i don't see any difference..

button is still there.

Ashley · ‎09-01-2021

Good Morning,

Can you try adding the "Isolate script" field to the UI Policy form and set it to false i.e. uncheck it.

Configure > Form Layout > Move field to the right of the bucket list

Try again.

It should be false to allow your DOM manipulation to work.

Kind Regards

Ashley

Prakhar7 · ‎11-12-2021

Even setting Isolate script to false, did not seem to work.

Jan Janou_ek · ‎05-10-2022

Hello,

I have the same problem, isolate script is false but button is still there. When I try to inject code directly to the page using JavaExecutor button disappears.

Jan Janou_ek · ‎05-16-2022

I had to revert conditions and it works now as expected.

if true:

function onCondition(){
$$('#show_test_flow_button')[0].show();
}

if false:

function onCondition() {
$$('#show_test_flow_button')[0].hide();
}