Need to return multiple values from script include.

[Swati44]

Hi All,

I need to return multiple values from a script include to onChange client script. I have tried to return object and an array but no luck. Please suggest how to achieve this.

Thanks,

Swathi

[Ahmmed Ali]

Hello,

please go through below link.

https://docs.servicenow.com/bundle/geneva-servicenow-platform/page/script/server_scripting/reference/r_ExamplesOfAsynchronousGlideAjax.html

Returning multiple values

Since the response is an XML document we are not limited to returning a single answer value. Here is a more complex example returning multiple XML nodes and attributes.

AJAX processor script include

/*
 * MyFavoritesAjax script include Description - sample AJAX processor returning multiple value pairs
 */ 
var MyFavoritesAjax = Class.create();
MyFavoritesAjax.prototype = Object.extendsObject(AbstractAjaxProcessor, {
 
	 /*
	 * method available to client scripts call using:
	 * var gajax = new GlideAjax("MyFavoritesAjax");
	 * gajax.addParam("sysparm_name","getFavorites");
	 */
	getFavorites: function() { // build new response xml element for result
         var result = this.newItem("result");
	  result.setAttribute("message","returning all favorites");
 
	  //add some favorite nodes with name and value attributes 
         this._addFavorite("color","blue");
         this._addFavorite("beer","lager");
         this._addFavorite("pet","dog");
       },
	 // all items are returned to the client through the inherited methods of AbstractAjaxProcessor
	 _addFavorite: function(name, value) { 
          var favs = this.newItem("favorite");
	   favs.setAttribute("name",name);
	   favs.setAttribute("value",value); },
 
	type:"MyFavoritesAjax"
 
 });

Client script

// new GlideAjax object referencing name of AJAX script include 
var ga = new GlideAjax("MyFavoritesAjax"); 
// add name parameter to define which function we want to call 
// method name in script include will be getFavorites
ga.addParam("sysparm_name","getFavorites");
 
// submit request to server, call ajaxResponse function with server response
 
ga.getXML(ajaxResponse);
 
function ajaxResponse(serverResponse) { 
  // get result element and attributes
  var result = serverResponse.responseXML.getElementsByTagName("result");
  var message = result[0].getAttribute("message");
 
  //check for message attribute and alert user
  if(message) alert(message);
 
  //build output to display on client for testing 
  var output = "";
 
  // get favorite elements 
  var favorites = serverResponse.responseXML.getElementsByTagName("favorite");
  for(var i = 0; i < favorites.length; i ++) { 
    var name = favorites[i].getAttribute("name");
    var value = favorites[i].getAttribute("value");
    output += name + " = " + value + "\n "; }
 
  alert(output); }

Thanks,

Ali

[Ankur Bawiskar]

Hi Swathi,

Following script should help you:

Script Include:

var obj = {};
obj.var1 = 'Hello';
obj.var2 = 'World';

return JSON.stringify(obj);

Client Script:

var obj = JSON.parse(answer);
alert(obj.var2);
alert(obj.var1);

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

[Devin Valencia2]

The simplest is often the best, thank you!

[manoj6699]

Hi Swathi,

if u want to return multiple sys_id's use return 'sys_idIN'+kk.sys_id.toString(); // Use toString or else it will give u only one record

else if u want to return multiple field values push all the different field values in to a string or an array

Method 1:

var sk;

if(sk==''){

sk = kk.fieldName;

}

else{

sk = sk+','+kk.fieldName;

}

Method 2:

var sk = [];

sk.push(sk.filedName.toString());//Use toString() or else it will show only one value