Reply To: Display image on condition of selection

• Jonathan

• June 13, 2015 at 7:20 am

Hi,

If you have a select field without multiple select it should just return a value not an array..

This code might get you in the right direction:

<?php 
$activity_rating = get_field('activity_rating');

switch($activity_rating){
	
	case '1':
		$img = 'urltoimage';
		break;
	case '2':
		$img = 'urltoimage2';
		break;
	case '3':
		$img = 'urltoimage3';
		break;
	case '4':
		$img = 'urltoimage4';
		break;
	case '5':
		$img = 'urltoimage5';
		break;
	case default:
		$img = 'fallbackimage';
		break;
		
	echo '<img src="' . $img . '" alt="" />';
	
}
?>