If only 1 field out of 4 is available - php

Question

I am using the below code on a WP site.
<?php
$images = array();
$images[] = $profile_user->banner_image_1;
$images[] = $profile_user->banner_image_2;
$images[] = $profile_user->banner_image_3;
$images[] = $profile_user->banner_image_4;
if(!empty($images[0]) || !empty($images[1])|| !empty($images[2])|| !empty($images[3])){
?>
<?php echo '<div class="slider2">'; ?>
<?php foreach($images as $img): ?>
<?php if(!empty($img)): ?>
<div>
<img src="
<?php
$image_id = $img;
$post_image_data = wp_get_attachment_image_src( $image_id, $size='profile_banner_img' );
echo $post_image_data[0];
?>" />
</div>
<?php endif; ?>
<?php endforeach; ?>
<?php echo '</div>'; ?>
<?php } ?>
What i need to do is use a "IF" or "array" statement or something that will see if only "banner_image_1" has been filled by user and display a different code.
In other words if banner_image_1 returns information but 2,3,4 dont i need remove the
<div class="slider2"> and the </div>

Answer 1

Try this:
<?php
$images = array();
for($i=1; $i<=4; $i++) {
if(!empty($profile_user->{"banner_image_$i"})) {
$images[] = $profile_user->{"banner_image_$i"};
}
}
$validPics = count($images);
if($validPics > 0) {
if($validPics > 1) echo '<div class="slider2">';
foreach($images as $img_id) {
$img_src = wp_get_attachment_image_src( $img_id, $size='profile_banner_img' );
echo '<div><img src=\"' . $img_src[0] . '" /></div>';
}
if($validPics > 1) echo '</div>';
}