How to inject a variable inside an SQL associated array? - sql

Question

I have many websites that use some of the same content snippets and instead of manually updating all the different websites, I thought it would be a good idea to have the content stored in a database as to only have one copy instead of multiple. It works great except for one issue which is the images that are in the article are sometimes left aligned and other times right aligned.
My solution was to add the following code to the article's image CSS tag that is in the database and use a variable on each of the individual pages to add the custom classes to the image.
class="<?php echo $ImgClass01; ?>"
EDIT: here is more of the content from what is stored in the database field to make my question a little more understandable.
<p><img src="img/charleston.jpg" class="<?php echo $ImgClass01; ?>">Is it the delightful year-round climate? The almost-European feel of its downtown city streets? The overwhelming...</p>
However, the webpage is only showing the text when viewing the source code and not using the variable. Almost anything is possible, but I'm not sure how to make this work.
Here is the code on the page...
// value for the class within the article to be printed on the page
$ImgClass01 = 'img-responsive img-rounded pull-right';
//Start a while loop to process all the rows
while ($row = mysql_fetch_assoc ($result_set))
{
$article = $row['article'];
echo $article;
} //END WHILE
EDIT: Just in case the entire page will be helpful, here is it.
<?php
$PageTitle = "Charleston, South Carolina | Local Towns";
$PageDescription = "Charleston is rated the first most popular vacation destination in the United States, and it surely must rank in...";
// 1. Create a database connection
$link = mysql_connect ("localhost", "root", "");
if (!$link) die("Could not connect: " . mysql_error());
// 2. Select a database to use
if (!mysql_select_db ("articleBank"))
die("Problem with the database: " . mysql_error());
// 3. Set up query for items to display
$query = "SELECT article FROM `articles` WHERE ID = 1";
// 4. Execute the query
$result_set = mysql_query ($query);
include ("theme/header.php");
// value for the class within the article to be printed on the page
$ImgClass01 = 'img-responsive img-rounded pull-right';
//Start a while loop to process all the rows
while ($row = mysql_fetch_assoc ($result_set))
{
$article = $row['article'];
echo $article;
} //END WHILE
// 5. Close Connection
mysql_close();
include ("theme/footer.php");
?>

Answer 1

while ($row = mysql_fetch_assoc ($other_result_set))
{
$ImgClass01 = $row['ImgClass01'];
}
not sure what you wanted, but you could use left, right in some row like place..