Display image on condition of selection

• pjsando

• April 29, 2015 at 7:14 pm

I would like to display different images depending on the value of a select dropdown. I have 5 options in the select dropdown (1 to 5) and a different image for each that I would like to display (image1 – image 5)
So far I have this code adapted from a previous post

<?php 
$activity_rating = get_field('activity_rating');
<?php if( is_array($activity_rating) && in_array( '1', $activity_rating ) ): ?>
<?php // the user has selected '1' select dropdown. Display image 1 ?>
<?php //Display image 1 ?>
<?php else if ( is_array($activity_rating) && in_array( '2', $activity_rating ) ): ?>
<?php //Display image 2 ?>
<?php else if ( is_array($activity_rating) && in_array( '3', $activity_rating ) ): ?>
<?php //Display image 3 ?>
<?php else if ( is_array($activity_rating) && in_array( '4', $activity_rating ) ): ?>
<?php //Display image 4 ?>
<?php else if ( is_array($activity_rating) && in_array( '5', $activity_rating ) ): ?>
<?php //Display image 5 ?>
<?php endif; ?>

Any ideas very much appreciate!
p

• 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="" />';
	
}
?>