PHP输出作为数据库的链接

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> <a href="../' . $row['description'] . '"> '. $row['title'] .'</a>';  
 }

}  

?>

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("<a href=\"%s\">%s</a>", $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.

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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.

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IMPORTANT: STOP using mysql_ functions because its deprecated from new PHP versions. Use PDO or mysqli instead.