Not getting variable

[Xtremer360]

What I'm want to know is either a better way to write this or find out what I'm doing because its not getting the shortName variable in the anchor tag.

 

function spotlight(){
$query = "
    SELECT 
        * 
    FROM 
        characters 
    WHERE 
        characters.statusID = 1 AND characters.styleID = 1 
    ORDER BY 
        RAND() 
    LIMIT 
        1"; 
$result = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($result);
    $shortName = $row[ 'shortName' ];
$labels = array('shortName');
    $img = array();
if($handle = opendir('images/spotlight/')) {
	$count = 0;
	while (false !== ($file = readdir($handle))) {
		if(strlen($file) > 2){
			$img[$count] = $file;
			$count++;
		}
	}
}
echo "<a href='bio.php?shortName='" . $shortName . "'><img src='/images/spotlight/".$img[rand(0, (count($img)-1))]."' alt=Spotlight border=0 /></a>";
} 

[wildteen88]

Remove the highlighted quote.

echo "<a href='bio.php?shortName='" . $shortName . "'><img src='/images/spotlight/".$img[rand(0, (count($img)-1))]."' alt=Spotlight border=0 /></a>";

This is ending the link too early and thus the shortname is not shown in the url.

 

Also a neater to way to do $img[rand(0, (count($img)-1))]. would be to use array_rand

[Xtremer360]

Well then if I remove that code how is it going to get turned into a link?

[wildteen88]

I meant the quotation mark, not the actual code I quoted. Sorry for not being clear.

 

I have edited my post to make it more clear.

[Xtremer360]

I apologize which quotation mark are you referring to?

[Xtremer360]

Thank you.

[fugix]

you can try this

echo "<a href='bio.php?shortName={$shortName}><img src='/images/spotlight/".$img[rand(0, (count($img)-1))]."' alt=Spotlight border=0 /></a>";
}