UI Action to create independent record ... ?

[Zod]

Hi Experts,

I'm using a UI action on a from to "copy" some of the fields and to new record that is to be created on a completely independent table.

Could anyone please help me with this?

Something like this I had in mind ... but missing the // CODE ??? part

****

function clickedReview() {

    gsftSubmit(null, g_form.getFormElement(), 'action_name');

        }

// Code that runs without 'onclick'

// Ensure call to server side function with no browser errors

if (typeof window == 'undefined')

    createNew();

function createnew() {

    // CODE ????

    current.update();

    gs.addInfoMessage(gs.getMessage("Record added"));

    action.setRedirectURL(current);

}

[shloke04 · Accepted solution]

Hi,

You can use the below Script in your UI Action

Script:

var x= new GlideRecord("change_request");     //Replace your Desired Target Table

x.initialize();

x.short_description = current.short_description;

x.assignment_group='Enter Sys Id';

x.insert();

action.setRedirectURL(current);

gs.addInfoMessage('Record <a href ="/Table_Name.do?sysparm_query=number%3D' + x.number + '">'+ x.number +'</a>'+ 'has been created.');

Similarly you can have other fields as it has been mentioned for Short Description and Assignment Group. In the above screenshot Replace Demand with your Table Name on which you want this UI Action to be present on the form.

Hope this helps.Mark the answer as correct/helpful based on impact.

Regards,

Shloke

[SanjivMeher]

Try below

ta = new GlideRecord('table_name');

ta.fieldname1= current.fieldname1;

ta.fieldname2= current.fieldname2;

ta.fieldname3= current.fieldname3;

ta.fieldname4= current.fieldname4;

ta.insert();

action.setRedirectURL(ta);