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hey my code wont show the else image

 

<?php $q = mysql_query($sql) or die("Error running query:".mysql_error());
                                 if($q && mysql_num_rows($q) > 0)  {
                                $row = mysql_fetch_array($q);
                                if(!empty($row)) {
                                echo "<img src='http://www.runningprofiles.com/members/images/thumbs/". $pid . "." . $row['ext']  . "'";
                                        }
                                        else {
                                    echo "<img src='http://www.runningprofiles.com/members/images/noimage.jpg>";
                                                }
																							                                            }
                                            ?>

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https://forums.phpfreaks.com/topic/91290-not-showing-else-statement/
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That's because $row will never be empty. If there are no records that code never goes inside the first IF loop. Also there is some duplicity in that code.:

 

Try this:

 

<?php

$q = mysql_query($sql) or die("Error running query:".mysql_error());

if($row = mysql_fetch_array($q))
{
    echo "<img src='http://www.runningprofiles.com/members/images/thumbs/". $pid . "." . $row['ext']  . "'";
}
else
{
    echo "<img src='http://www.runningprofiles.com/members/images/noimage.jpg>";
}

?>

 

Edit:

 

Since you do this

if($q && mysql_num_rows($q) > 0)  {

 

This is kind of irrelevant

if(!empty($row)) {

hey i dont belive i finished the code off properly because all i get for an image is

http://www.runningprofiles.com/members/images/noimage.jpg%3E%20%20%20%20 ... ect ect

<?php 

$q = mysql_query($sql) or die("Error running query:".mysql_error());

if($row = mysql_fetch_array($q))
{
    echo "<img src='http://www.runningprofiles.com/members/images/thumbs/". $pid . "." . $row['ext']  . "'";
}
else
{
    echo "<img src='http://www.runningprofiles.com/members/images/noimage.jpg";}

?>

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