爲什麼心不是我的PHP代碼更新SQL

有誰知道爲什麼PHP代碼沒有更新列pictures將更新他們的休息，但只是沒有圖片列是更新用戶信息，以便電子郵件地址，密碼和圖片我是很新的PHP，所以我真的不知道要尋找什麼，當即時尋找錯誤爲什麼心不是我的PHP代碼更新SQL

<?php 

    require("common.php"); 

    if(empty($_SESSION['user'])) 
    { 
     header("Location: login.php"); 

     die("Redirecting to login.php"); 
    } 

    if(!empty($_POST)) 
    { 
     if(!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) 
     { 
      die("Invalid E-Mail Address"); 
     } 

     if($_POST['email'] != $_SESSION['user']['email']['picture']) 
     { 
      $query = " 
       SELECT 
        1 
       FROM users 
       WHERE 
        email = :email 
        picture = :picture 
      "; 

      $query_params = array( 
       ':email' => $_POST['email'] 
      ); 

      try 
      { 
       $stmt = $db->prepare($query); 
       $result = $stmt->execute($query_params); 
      } 
      catch(PDOException $ex) 
      { 
       die("Failed to run query: " . $ex->getMessage()); 
      } 

      $row = $stmt->fetch(); 
      if($row) 
      { 
       die("This E-Mail address is already in use"); 
      } 
     } 

     if(!empty($_POST['password'])) 
     { 
      $salt = dechex(mt_rand(0, 2147483647)) . dechex(mt_rand(0, 2147483647)); 
      $password = hash('sha256', $_POST['password'] . $salt); 
      for($round = 0; $round < 65536; $round++) 
      { 
       $password = hash('sha256', $password . $salt); 
      } 
     } 
     else 
     { 
      $password = null; 
      $salt = null; 
     } 

     $query_params = array( 
      ':email' => $_POST['email'], 
      ':user_id' => $_SESSION['user']['id'], 
      ':picture' => $_POST['picture'], 
     ); 

     if($password !== null) 
     { 
      $query_params[':password'] = $password; 
      $query_params[':salt'] = $salt; 
     } 

     $query = " 
      UPDATE users 
      SET 
       email = :email 
       picture = :picture 
     "; 

     if($password !== null) 
     { 
      $query .= " 
       , password = :password 
       , salt = :salt 
      "; 
     } 

     $query .= " 
      WHERE 
       id = :user_id 
     "; 

     try 
     { 
      // Execute the query 
      $stmt = $db->prepare($query); 
      $result = $stmt->execute($query_params); 
     } 
     catch(PDOException $ex) 
     { 
      die("Failed to run query: " . $ex->getMessage()); 
     } 

     $_SESSION['user']['email']['picture'] = $_POST['email']; 

     header("Location: private.php"); 

     die("Redirecting to private.php"); 
    } 

?>

來源

2013-12-18 user3116240

+1的實際使用準備好的語句和驗證輸入 –

您需要添加'電子郵件=之間的逗號：電子郵件圖片=：picture'。你正在檢查'$ _POST ['picture']'是否被設置？ – andrewsi

還需要在SELECT的WHERE子句中使用AND，否？ –

你失蹤後一個逗號：電子郵件在您的UPDATE語句。

$query = " 
      UPDATE users 
      SET 
       email = :email 
       picture = :picture 
     ";

應該

$query = " 
     UPDATE users 
     SET 
      email = :email, 
      picture = :picture 
    ";

編輯：除了這個，你也缺少在第一個查詢參數：

 $query = " 
      SELECT 
       1 
      FROM users 
      WHERE 
       email = :email 
       picture = :picture 
     "; 

     $query_params = array( 
      ':email' => $_POST['email'] 
     );

通知你如何只應用：電子郵件你查詢參數，但您的查詢需要：email和：圖片。

你要麼需要從$query刪除picture = :picture或添加':picture' => $_POST['picture']到$query_params

來源

2013-12-18 18:47:18 Matt

好吧，我剛剛更新它，它還是老樣子dosnt似乎工作 – user3116240

檢查以確保你的表單輸入有'NAME =「圖片」'，拼寫正確。你確定'picture'不是一個文件嗎？ – Leng

在$ query語句之前嘗試var_dump（$ _ POST ['picture']）並查看該值是什麼。 – Matt

你缺少一個,。

$query = " 
    UPDATE users 
    SET 
     email = :email 
     picture = :picture 
";

您需要將其更改爲

$query = " 
    UPDATE users 
    SET 
     email = :email, 
     picture = :picture 
";

來源

2013-12-18 18:47:22

爲什麼心不是我的PHP代碼更新SQL

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