Show UI Action form button during choice field on change

[Andrew48]

Hello Snowers,

I have a UI Action form button "Send Notification", which has to be shown only when a certain value is chosen from the status choice list. 

I have provided the condition but it is working only when the form is loading. is there a way to make this happen without saving the form and reloading it.

The UI Action i configured.

Also a clarification, does the UI action form button gets executed only during a form load?

[Priyanka136]

Hi Andrew,

Please refer the below link it might be helpful:-

https://community.servicenow.com/community?id=community_question&sys_id=2cf54361db1cdbc01dcaf3231f96...

Let me know if you have any question !!!

Thanks,
Priyanka

[Andrew48]

Hi Priya,

Thanks for your response.

I tried using the same code in a client script, as there was a onchange client script present already, 

Even after i checked the isolate script check box, i am facing the below error.

TypeError: $$ is not a function function () { [native code] }

I used this code "$$('send_notification')[0].show(); "

kindly help.

[Priyanka136]

Hey Andrew,

Make "Isolate" checkbox to false then try to execute.

Thanks,

Priyanka

[Andrew48]

Hello,

Thanks for your response.

I tried using the same code in a client script, as there was a onchange client script present already, 

Even after i checked the isolate script check box, i am facing the below error.

TypeError: $$ is not a function function () { [native code] }

I used this code "$$('send_notification')[0].show(); "

kindly help.

[Ankur Bawiskar]

Hi Andrew,

Isolate script field should be set to false;

Also can you try

$('send_notification')[0].show();

Mark ✅ Correct if this solves your issue and also mark 👍 Helpful if you find my response worthy based on the impact.
Thanks
Ankur