different div is result has value more than 1

[lovephp]

how do i do this someone?

 

here is my code

 

what is want is if there is just just one image then it display as the code below

 

              <?php
              $stm = $db->prepare("SELECT * FROM images WHERE id = '".$id."'");
              $stm->execute();
              $rs = $stm->fetchAll();
              foreach( $rs as $img ) {
              ?>
                <div style="display: inline-block;">
                 <img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/>
                </div>
              <?php
                }                        
               ?>  

 

if more than 1 record than only the first image will be in

 

<div style="display: inline-block;"><img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/></div>

and rest will be inside just

 

<div><img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/> </div>
<div><img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/> </div>
<div><img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/> </div>
<div><img src="uploads/images/<?php echo $img['img']; ?>" class="fancybox"/> </div>

 

 

without the style="display: inline-block;" from 2nd record onwards

[requinix]

counter = 1

for each image {

	if counter = 1 {

		output div code with inline-block

	} else {

		output plain div code

	}

	output image code

	output closing div


	counter++

}

[lovephp]

not getting it bro

[lovephp]

counter = 1
for each image {
	if counter = 1 {
		output div code with inline-block
	} else {
		output plain div code
	}
	output image code
	output closing div

	counter++
}

still not getting it how to get it into my codes

[lovephp]

did it thanks a ton

[Jacques1]

We have CSS for this: :first-child