[RESOLVED] Will not display results correctly, need help

[Seraphino]

I have been working on a script long enough to decide that I am now officially frustrated. For some reason this chunk of the script will not display things properly...either it leaves off results, does not display proper URLs, or it leaves off links.

I don't know why and I can not figure it out, if anyone can help that'd be greatly appreciated.

elseif ($do == "list")
{
    $artist_name = $_POST['artist_name'];
    $query = "SELECT * FROM Assigns WHERE Username='" . $artist_name . "'";    
    $resulta = mysql_query($query);
    $num = mysql_num_rows($result);
    $xn = 0;
    
    while ($x = mysql_fetch_array($resulta))
    {    
        echo "<strong>" . $x['File Name'] . "</strong><br /><a href=" . "uploads/" . $artist_name . "/" . $x['Directory'] . "/" . $x['File Name'] . ">Download This Model</a>";
        
        if ($x['HasImage'] == yes)
        {
            $query_images = "SELECT * FROM Images WHERE Username='" . $artist_name . "' AND Directory='" . mysql_result($result, $x, "Directory") . "'";
            $result_images = mysql_query($query_images);
            
            echo "<br /><a href=" . "uploads/" . $artist_name . "/" . $x['Directory'] . "/" . mysql_result($result_images, $xn, "Image Assigned") . ">View Image</a><hr>";
        }
        else
        {
            echo "<hr>";
        }
        
        $x++;
        $xn++;
    }
} 

[visualAd]

What does it do? And why is the echo statement being used to output HTML? - that is a slapable offence.

[Seraphino]

This creates a dynamic html page where artists (on a game dev team) can download each other's models and view screenshots of the models they are downloading.

Also, the reason it is outputting HTML is becuase I didn't want to deal with parenthesis at the time. It will probably be changed.

This page is not really open to the public, just to certain people.

[kows]

the reason you aren't getting proper URLs is most likely because you're using terrible markup.

considering you're sloppy enough to use spaces in a table's field names (or at least, I personally consider that sloppy!), I assume your $artist_name variable might also contain spaces, along with your directory -- and maybe even the file name. if you're creating a hyperlink, you must enclose strings that have spaces in them with quotations; otherwise, the browser doesn't know if you're trying to continue your URL, or add more parameters. if you have no idea what I mean, then take a quick look at this example:

<a href="http://domain.com/this is a link with spaces/file name with spaces.gif">this is a link with spaces -- it will work correctly.</a>

/*
  the above will link to the following URL, just like you wanted:
  http://domain.com/this is a link with spaces/file name with spaces.gif
*/

<a href=http://domain.com/this is a link with spaces/file name with spaces.gif>this is a link with spaces. it will NOT work!</a>

/*
  the above will link to the following URL, which is NOT what you wanted:
  http://domain.com/this
*/ 

quick edit: why are you incrementing "$x"? $x is a database result. you can't increment it.

[Seraphino]

All my problems were solved, thanks for helping you guys!