profilid- siger Profilen med den angivne id, findes desværre ikke

Er det okay jeg vise dig det på teamwiever?

du kan skrive til mig på djwillumsupport@gmail.com

PHP kode 

 if($num_rows == null) {

PHP kode 

 if($num_rows == 0) {

PHP kode 

 if($num_rows == 0) {
   echo "Profilen med den angivne id, findes desværre ikke.";
 } else {
   $row3 = mysql_fetch_assoc($select);
 
 $rank = mysql_query("SELECT * FROM brugere INNER JOIN rank ON brugere.rank = rank.id ORDER BY '". $_GET['id'] ."' ");
 $rrow = mysql_fetch_assoc($rank);

PHP kode 

 <?php
   $select = mysql_query("SELECT * FROM brugere WHERE id = '". $_GET['id'] ."' ");
   $num_rows = mysql_num_rows($select);
 
   $select_user = mysql_query("SELECT * FROM brugere WHERE id = '". $_SESSION['id'] ."' ");
   $use = mysql_fetch_assoc($select_user);
 
   $select2 = mysql_query("SELECT * FROM p_opslag WHERE bid = '". $_GET['id'] ."' ORDER BY id DESC LIMIT 4");
   $num_rows2 = mysql_num_rows($select2);
 
   $select3 = mysql_query("SELECT * FROM p_opslag WHERE bid = '". $_GET['id'] ."' && bid2 = '". $_GET['id'] ."' ORDER BY id DESC LIMIT 10 ");
   $oprow = mysql_fetch_assoc($select3);
   $num_rows_opslag = mysql_num_rows($select3);
 
 
   if(isset($_POST['send'])) {
     mysql_query("INSERT INTO p_opslag (bid, bid2, indhold, ip) VALUES ('". $_GET['id'] ."', '". $_POST['bid2'] ."', '". $_POST['indhold'] ."', '". $_POST['ip'] ."')");
     header("Location: http://www.b2f.dk/?s=profile&id=". $_GET['id'] ."");
   }
 
   if($_GET['d']=="true" && $_GET['s']=="profile" && $_SESSION['id']==$_GET['id']) {
     mysql_query("DELETE FROM p_opslag WHERE id = '". $_GET['pid'] ."'");
     header("Location: http://www.b2f.dk/?s=profile&id=". $_GET['id']."");
   }
 
 function CalculateAge($BirthDate)
 {
         list($Year, $Month, $Day) = explode("/", $BirthDate);
  
         $YearDiff = date("Y") - $Year;
  
         if(date("m") < $Month || (date("m") == $Month && date("d") < $DayDiff))
         {
                 $YearDiff--;
         }
         return $YearDiff;
 }
 
 if($num_rows == 0) {
   echo "Profilen med den angivne id, findes desværre ikke.";
 } else {
   $row3 = mysql_fetch_assoc($select);
 
 $rank = mysql_query("SELECT * FROM brugere INNER JOIN rank ON brugere.rank = rank.id ORDER BY '". $_GET['id'] ."' ");
 $rrow = mysql_fetch_assoc($rank);
 
 if($row3['img'] == null) {
   $row3['img'] = "http://www.fairtrade-denmark.dk/fileadmin/templates/Billeder/Fairtrade/no_image.gif";
 }
 
 if($row3['fdag'] == null) {
   $row3['fdag'] = date("d");
   $row3['fmaaned'] = date("m");
   $row3['faar'] = date("Y");
 }
 
 ?>
 <script type="text/javascript" language="JavaScript">
 function ReverseContentDisplay(d) {
 if(d.length < 1) { return; }
 if(document.getElementById(d).style.display == "block") { document.getElementById(d).style.display = "none"; }
 else { document.getElementById(d).style.display = "block"; }
 }
 
 document.getElementById(d).style.display = "none";
 </script>
 
 
 <h3><?php echo $row3['brugernavn']; ?>'s profil</h3>
 <div style="color: #999999; font-size: 10px;">
 <?php if($row3['rank']=="1") { ?>
 Denne profil er ægte
 <?php } ?>
 </div>
 
 <div style="float: left; border-right: 1px dotted #999999; width: 220px; min-height: 300px;">
 <img src="<?php echo $row3['img']; ?>" border="2" height="200" width="200" />
 <strong>Navn:</strong> <?php echo $row3['fnavn'] . " " . $row3['enavn']; ?><br />
 <strong>Alder:</strong> <?php echo CalculateAge($row3['faar'] . "/" . $row3['fmaaned'] . "/" . $row['fdag']); ?> år<br />
 <strong>Rank:</strong> <?php echo $rrow['nrrank']; ?>
 </div>
 <?php } ?>

Lige efter $select = mysql_query("SELECT * FROM brugere WHERE id = '". $_GET['id'] ."' ");
f.eks.

jamen hvad skulle jeg skrive ?? og hvor henne ved $select = mysql_query("SELECT * FROM brugere WHERE id = '". $_GET['id'] ."' ");

jeg synes jeg kan se flere steder i starten.

PHP kode 

 <?php
    $select = mysql_query("SELECT * FROM brugere WHERE id = '". $_GET['id'] ."' ");
    $num_rows = mysql_num_rows($select);
    var_dump($select)
 
 .......