PHP登錄腳本不給預期的結果，或任何

我有一個用簡單的形式訪問的PHP登錄腳本：PHP登錄腳本不給預期的結果，或任何

<?php 
session_start(); 
try{ 
    $user = 'root'; 
    $pass = null; 
    $pdo = new PDO('mysql:host=localhost; dbname=divebay;', $user, $pass); 

    if(isset($_SESSION['loggedin'])){ 

     echo "1"; //already logged in 
    } 


    else{ 
     $username = $_POST['username']; 
     $password = sha1($_POST['password']); 

     $ucheck = $pdo->prepare('SELECT * FROM user WHERE username = ?'); 
     $ucheck->bindValue(1, $username); 
     $ucheck->execute(); 

     if($ucheck->fetch(PDO::FETCH_OBJ)){ 
      $stmt = $pdo->prepare('SELECT * FROM user WHERE username = ? AND password = ?'); 
      $stmt->bindValue(1, $username); 
      $stmt->bindValue(2, $password); 

      if($stmt->fetch(PDO::FETCH_OBJ)){ 
       $row = $stmt->fetch(PDO::FETCH_ASSOC); 

       $_SESSION['username'] = $row['username']; 
       $_SESSION['loggedin'] = 'YES'; 
       $_SESSION['location'] = $row['location']; 
       echo "2"; //logged in 

      } 
      else{ 
       echo "3"; //password incorrect 
      } 
     } 
     else{ 
      echo "4"; //user does not exist 
     } 

    } 
}catch(PDOException $e){ 
    echo $e->getMessage(); 
} 

?>

，但是當我嘗試使用我剛剛創建的帳戶來運行它，並有證實存在於數據庫中，我沒有從這個腳本得到任何迴應。我希望它回聲2鑑於登錄信息是正確的，但我什麼也沒得到

任何人都可以提出什麼ive做錯了嗎？

來源

2012-09-05 Bundy

你可以把'error_reporting（E_ALL）; ini_set（「display_errors」，1）;'在腳本的頂部，看看它是否顯示任何東西？ – bretterer

我會把我的錢放在'$ pdo'爲false或null而不是對象:) –

看起來你忘了聲明：

if($ucheck->fetch(PDO::FETCH_OBJ)){ 
     $stmt = $pdo->prepare('SELECT * FROM user WHERE username = ? AND password = ?'); 
     $stmt->bindValue(1, $username); 
     $stmt->bindValue(2, $password); 

     // Execute it!!! 
     if ($stmt->execute()) { 
      $row = $stmt->fetch(PDO::FETCH_OBJ); 

      if ($row) { 
      // And don't call fetch() again, since you would already have advanced 
      // the record pointer in the first fetch() above. If one record was returned, 
      // this one would always be FALSE. 
      //$row = $stmt->fetch(PDO::FETCH_ASSOC); 

      $_SESSION['username'] = $row['username']; 
      $_SESSION['loggedin'] = 'YES'; 
      $_SESSION['location'] = $row['location']; 
      echo "2"; //logged in 
      } 
      // else execute failed... 
     }

來源

2012-09-05 01:57:35

雖然沒有'execute（）'它應該仍然顯示'3'，因爲結果是空的。 –

你肯定session.use_cookies = 1在php.ini？

請確保名稱是PHPSESSION cookie。

來源

2012-09-05 02:17:51 SamPeng

PHP登錄腳本不給預期的結果，或任何

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