PHP output as link from database - php

Question

I have the following code that works by outputting as a link ( the link comes from a field in my database) I wish to do the same for the code below, however i cannot get it work, here is the example of what I have that works, and the code that i wish to make output as a link:
Working Code what I want it to look like
if (!empty($_REQUEST['term'])) {
$term = mysql_real_escape_string($_REQUEST['term']);
$sql = "SELECT * FROM adrenaline WHERE title LIKE '%".$term."%'";
$r_query = mysql_query($sql);
while ($row = mysql_fetch_array($r_query)){
echo '<br> '. $row['title'] .'';
}
}
?>
And the code that i have at the moment, it works by be manually typing in the hyper link, however I wish to make it take the link from the database like the example above
//query the database
$query = mysql_query("SELECT * FROM hobby WHERE id = '1' ");
//ferch the results / convert results into an array
WHILE($rows = mysql_fetch_array($query)):
$title = $rows['title'];
echo "<a href='shard.php'>$title</a>";
endwhile;
?>
Many thanks!

Answer 1

I am not 100% certain if this is what you meant to ask... let me know in comments:
<?PHP
$query = mysql_query("SELECT * FROM hobby WHERE id = '1' ");
if(mysql_num_rows($query) >= 1) {
while($rows = mysql_fetch_array($query)) {
echo sprintf("%s", $rows["description"], $rows["title"]);
}
} else { echo "No hobbies found."; }
?>
I believe you might have faced some syntax issues while dealing with quotes parsing a variable in <a html tag. Consider using sprintf something like in my example.
I have also added a mysql_num_rows() just in case and you can see its a good fail-safe method incase there are no rews found on any select query.
IMPORTANT: STOP using mysql_ functions because its deprecated from new PHP versions. Use PDO or mysqli instead.