Home › Forums › General Issues › Display image on condition of selection › Reply To: Display image on condition of selection
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="" />';
}
?>
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