hiding content when variable is not present - php

Question

I am using php to publish news articles from my mysql database. The articles have images placed within their text. The images are stored as variables, each its own path. The problem is I don't always use the full complement of images for each article, and the way I've written it, a broken image displays.
In short, how can I keep broken images from showing up if say '$image_ii' or '$image_iii' does not exist for a specific ID.
The way I have it written out is as follows:
<?php
include ('connect.php');
$friendly_url=$_GET['friendly_url'];
$query = "SELECT * FROM entries WHERE friendly_url='$friendly_url'";
$entries = mysql_query($query);
while($row = mysql_fetch_array($entries, MYSQL_ASSOC))
{
$title = $row['title'];
$author = $row['author'];
$pub_date = $row['pub_date'];
$content_i = $row['content_i'];
$content_ii = $row['content_ii'];
$content_iii = $row['content_iii'];
$content_iv = $row['content_iv'];
$image_i = $row['image_i'];
$image_ii = $row['image_ii'];
$image_iii = $row['image_iii'];
$image_caption_i = $row['image_caption_i'];
$image_caption_ii = $row['image_caption_ii'];
$image_caption_iii = $row['image_caption_iii'];
$pre_link = $row['pre_link'];
$post_link = $row ['post-link'];
$id = $row['id'];
$friendly_url = $row['friendly_url'];
$american_date = date("F d, Y", strtotime($pub_date));
$friendly_content_i = nl2br($content_i);
$friendly_content_ii = nl2br($content_ii);
$friendly_content_iii = nl2br($content_iii);
$friendly_content_iv = nl2br($content_iv);
echo "
<div id='post'>
<div id='post-title'>$title</div>
<div id='post-content'
$friendly_content_i
<div id='media'> <img src='../$image_i' class='scale-image'> </div>
<div id='media-caption'> $image_caption_i </div>
$friendly_content_ii
<div id='media'> <img src='../$image_ii' class='scale-image'> </div>
<div id='media-caption'> $image_caption_ii </div>
$friendly_content_iii
<div id='media'> <img src='../$image_iii' class='scale-image'> </div>
<div id='media-caption'> $image_caption_iii </div>
$friendly_content_iv
</div>
<div id='post-footer'> by $author <br> $american_date </div>
</div>
</div> ";
} mysql_close($connection); ?>
Apologies if this is terrible, it's my first dynamic website.

Answer 1

Try using the following snippet
if (file_exists("../".$image_iii)) {
echo "<img src='../$image_iii' class='scale-image'>";
} else {
echo "<img src='../default-image' class='scale-image'>";
}

Answer 2

I ended up finding a solution that worked using:
if (!empty($image_iii)) { echo "<img src='../$image_iii' class='scale-image'>"; }
else { echo "";}
That said, clearly you kind folks are right about me needing to swap over to mysqli...found this website as a good resource for migrating code to mysqli. Thanks to everyone who responded, everyones comments were helpful.

Answer 3

You already told us the solution to your problem:
if say '$image_ii' or '$image_iii' does not exist for a specific ID.
Use if file_exists() to check that your image exists, and don't output it, if it doesn't.
Furthermore I see you're using mysql_query - This is deprecated and very unsafe. You should use mysqli or PDO instead