How to know the field name which is set display as true in table?

[Community Alums]

I have a table X, in that i have three fields as a,b,c.

if i reference this table i have set 'a' field as display to true.

if i give the table name i want to get field name of which i was set display to true.

can any one help me in this issue.

[Community Alums]

Thanks paul,

but sys_user is solo class right? there is no base table for sys_user.

in that case field name not getting.

Thanks,

[arnabwa]

Hi Rajendra,

Go to the concerned table >> Open >> See the "Table Columns" area >> You will see a column named "Display" >> Here you will find a particular row being kept as true. This is the row that is available when we reference it from a form. If you do not see the Display field, Click on the gear icon and select it from the Slush bucket.

Hope you got it.

Thanks,

Arnab

[Geoffrey2 · Accepted solution]

Try this code.   It does a complete search. It worked on all the tables I tested.

function getDisplayField(table) {
       var i, currentTable, deo, dict;

       var hierarchy = String(new TableUtils(table).getHierarchy());
       hierarchy = hierarchy.substring(1,hierarchy.length-1).split(', '); // turn List into Array

       for (i = 0; i < hierarchy.length; i++) {
               currentTable = hierarchy[i];

               // Check for Dictionary Entry Overrides first
               deo = new GlideRecord('sys_dictionary_override');
               deo.addQuery('name', currentTable);
               deo.addQuery('display_override', true);
               deo.query();
               if (deo.next())
                       return String(deo.element);

               // Check the Dictionary next
               dict = new GlideRecord('sys_dictionary');
               dict.addQuery('name', currentTable);
               dict.addQuery('display', true); 
               dict.query();
               if (dict.next())
                       return String(dict.element);
       }

       // No display values have been defined.   Look for a default field
       var defaults = ['number','u_number','name','u_name'];
       for (i = 0; i < defaults.length; i++) {
               dict = new GlideRecord('sys_dictionary');
               dict.addQuery('name', 'IN', hierarchy);
               dict.addQuery('element', defaults[i]);
               dict.query();
               if (dict.next())
                       return String(defaults[i]);
       }

       // No display value
       return 'sys_id';
}
var displayField = getDisplayField('sys_user');
gs.log('displayField: ' + displayField);

[Marek Meres]

Thanks - great script, but... I ran it for 'cmdb_ci_server' and it returned:

*** Script: displayField: u_number

I modified the script a bit adding the table the attribute was found in and it returned:

*** Script: displayField: u_number, defined in table: cmdb_ci_tape_server

So I believe it should use TableUtils(table).getTables() and not TableUtils(table).getHierarchy() - this way only the "parent" path is checked. With this modification the function returned (correct imho)

*** Script: displayField: name, defined in table: cmdb

So finally:

// from: https://community.servicenow.com/community?id=community_question&sys_id=e18e4b21dbdcdbc01dcaf3231f9619f8
// credits to: Geoffrey (https://community.servicenow.com/community?id=community_user_profile&user=43521e29dbd81fc09c9ffb651f961955)


function getDisplayField(table) {
	// MM: added table to results so returning object
	var result = {}
	
	var i, currentTable, deo, dict;
	// MM: getting hierarchy includes child tables - which is wrong! should be getTables (only parents)
	//var hierarchy = String(new TableUtils(table).getHierarchy());
	var hierarchy = String(new TableUtils(table).getTables());
	hierarchy = hierarchy.substring(1,hierarchy.length-1).split(', '); // turn List into Array
	for (i = 0; i < hierarchy.length; i++) {
		currentTable = hierarchy[i];
		// Check for Dictionary Entry Overrides first
		deo = new GlideRecord('sys_dictionary_override');
		deo.addQuery('name', currentTable);
		deo.addQuery('display_override', true);
		deo.query();
		if (deo.next())
		{
			// MM: added table to results so returning object
			result.table = currentTable;
			result.field = String(deo.element);
			return result;
		}
		// Check the Dictionary next
		dict = new GlideRecord('sys_dictionary');
		dict.addQuery('name', currentTable);
		dict.addQuery('display', true); 
		dict.query();
		if (dict.next())
		{
			// MM: added table to results so returning object
			result.table = currentTable;
			result.field = String(dict.element);
			return result;
		}			
	}
	// No display values have been defined.   Look for a default field
	var defaults = ['number','u_number','name','u_name'];
	for (i = 0; i < defaults.length; i++) {
		dict = new GlideRecord('sys_dictionary');
		dict.addQuery('name', 'IN', hierarchy);
		dict.addQuery('element', defaults[i]);
		dict.query();
		if (dict.next())
		{
			// MM: added table to results so returning object
			result.table = currentTable;
			result.field = String(defaults[i]);
			return result;
		}				
	}
	// No display value
	// MM: added table to results so returning object
	result.table = currentTable;
	result.field = 'sys_id';	
	return result;
}

var displayField = getDisplayField('sys_user');

gs.log('displayField: ' + displayField.field + ", defined in table: "+displayField.table);

[Kieran Anson]

For those that find this in the future from Google searches. I've created an article that provides a condensed version now that there is a isDisplay() function.

Easily Retrieve a Table Display Field Name