dongshi1606 2010-07-20 19:02
浏览 48

论坛类别问题

i have a forum and in that forum people can create a thread in a category they chose, but i wanna check if the category exists but i dont know how to, because know if i type another category number in the URL i still see the page.

here is my code

    $kategoriID = $_GET['kategoriID'];

if(!isset($overskrift) || !isset($indhold) || !isset($fejl))
{
 $overskrift = "";
 $indhold    = "";
 $fejl   = false;


}#Lukker if isset overskrift,indhold


 if(isset($kategoriID))
 {

   #tjek om kategori findes
  $mysql = connect();
        $stmt = $mysql->prepare("SELECT count(fk_forum_kategori) As t_id FROM forum_traad WHERE fk_forum_kategori = '$kategoriID'") or die($mysql->error);
        $stmt->bind_result($t_id);
        $stmt->execute() or die($mysql->error);
  $stmt->fetch();

  if($t_id <= 0)
  {
   echo $t_id;
   $fejl == true;
   echo "<br>HEj med dig!";
   exit();
  } 
 }#Lukker isset







if(isset($_POST['send'])) {

 $kategoriID = $_GET['kategoriID'];
 $overskrift = htmlspecialchars($_POST['overskrift']);
 $indhold    = htmlspecialchars($_POST['indhold']);
 $godkendt   = "ja";

 if($fejl == true)
 {
  $error = "Denne kategori findes ikke"; 
 } elseif(empty($overskrift) || empty($indhold)) {
  $error = "Alle felter skal udfyldes";
 } else {
  $mysql = connect();
  $stmt = $mysql->prepare("INSERT INTO forum_traad (overskrift, indhold, fk_forum_kategori, brugernavn, dato, godkendt) VALUES (?,?,?,?,?,?)") or die($mysql->error);
  $stmt->bind_param('ssisis', $overskrift, $indhold, $kategoriID, $_SESSION['username'], $dato, $godkendt) or die($mysql->error);
  $stmt->execute();
  $stmt->close();

  $traadID = mysqli_insert_id($mysql);

  header("location: forum.traad.php?traadID=$traadID&kategoriID=$kategoriID");
 }#Lukker else




}#Lukker isset send
  • 写回答

2条回答 默认 最新

  • dor2p0520 2010-07-20 19:15
    关注
    "SELECT count(fk_forum_kategori) As t_id 
    

    Probably isn't what you want here. Using count() will return the number of rows that match the criteria following it in your SQL query, so it's likely to always show you category number 1 (since each category number will only appear once in the database).

    You will probably want something like this instead:

    "SELECT fk_forum_kategori FROM forum_traad WHERE fk_forum_kategori = '$kategoriID'"
    
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