一個數據庫的作品，第二不

我有一個數據庫，在它有4個表，2個私人用戶和2個業務用戶。由於某種原因，當我嘗試使用商業用戶的電子郵件登錄時，它不起作用，但用戶名可以正常工作，並且在私人表格中起作用，這裏是我的代碼，如果我沒有正確解釋它，告訴我和虐待盡我所能解釋再次一個數據庫的作品，第二不

$password = $_POST['password']; 
    $emailuser = $_POST['unameemail']; 
    $password = mysqli_real_escape_string($sql , $password); 
    $emailuser = mysqli_real_escape_string($sql , $emailuser); 
    $pwcheck = " 
    SELECT * FROM private AS p 
    INNER JOIN user_private_data 
    AS c ON p.id = c.id 
    WHERE username='$emailuser' OR email='$emailuser'"; // part that works fine 
    $resultcheck = mysqli_query($sql , $pwcheck); // part that works fine 
    $rowcheck = mysqli_fetch_array($resultcheck , MYSQLI_ASSOC); // part that works fine 
    $hash = $rowcheck['password']; // part that works fine 
    $hash_pwd = password_verify($password , $hash); 
    if ($hash_pwd != 0) { 
     $_SESSION['username'] = $rowcheck['username']; // part that works fine 
     $_SESSION['logged'] = true; // part that works fine 
     header("refresh:0;url=../blablabla.php");  // part that works fine     
    } else { 
     $privateuser = " 
     SELECT * FROM business AS d 
     INNER JOIN user_business_data 
     AS j ON d.id = j.id 
     WHERE username='$emailuser' OR email='$emailuser'"; // doesn't work 
     $resultprivate = mysqli_query($sql , $privateuser); // doesn't work 
      $rowprivate = mysqli_fetch_array($resultprivate , MYSQLI_ASSOC); 
     $hashprivate = $rowprivate['password']; 
     $hash_private = password_verify($password , $hashprivate); 
     if ($hash_private != 0) { 

      $_SESSION['username'] = $rowprivate['username']; 
      $_SESSION['logged'] = true; 
      $_SESSION['business'] = $rowprivate['bname']; 
      $_SESSION['type'] = 'business'; 
}

來源

2017-02-07 moran

請在解析之前，淨化你的輸入值進入您的查詢。這個腳本很容易受到SQL注入的影響。 – Akintunde007

你的腳本存在[SQL注入攻擊]的風險（http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php）看看發生了什麼事[小鮑比表]（http://bobby-tables.com/）即使 [如果你是逃避投入，它不安全！]（http://stackoverflow.com/questions/5741187/sql-injection-that-gets -around-mysql-real-escape-string）使用[prepared parameterized statements]（http://php.net/manual/en/mysqli.quickstart.prepared-statements.php） – RiggsFolly

這是登錄頁面而不是註冊表頁面 – moran

試試這個：你需要檢查，如果該查詢實際上有取前一個結果，我假設變量$ SQL是在您的連接定義

<?php 

$password = $_POST['password']; 
$emailuser = $_POST['unameemail']; 
$password = mysqli_real_escape_string($sql, $password); 
$emailuser = mysqli_real_escape_string($sql, $emailuser); 
$pwcheck = " 
      SELECT * FROM private AS p 
      INNER JOIN user_private_data 
      AS c ON p.id = c.id 
      WHERE username='$emailuser' OR email='$emailuser'"; // part that works fine 
$resultcheck = mysqli_query($sql, $pwcheck); // part that works fine 
$rowcheck = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC); // part that works fine 
$hash = $rowcheck['password']; // part that works fine 
$hash_pwd = password_verify($password, $hash); 
if ($hash_pwd != 0) { 
    $_SESSION['username'] = $rowcheck['username']; // part that works fine 
    $_SESSION['logged'] = true; // part that works fine 
    header("refresh:0;url=../blablabla.php");  // part that works fine     
} else { 
    $privateuser = " 
       SELECT * FROM business AS d 
       INNER JOIN user_business_data 
       AS j ON d.id = j.id 
       WHERE username='$emailuser' OR email='$emailuser'"; // doesn't work 
    $resultprivate = mysqli_query($sql, $privateuser); // doesn't work 
    if ($resultprivate->num_rows > 0) { 
     $rowprivate = mysqli_fetch_array($resultprivate, MYSQLI_ASSOC); 
     $hashprivate = $rowprivate['password']; 
     $hash_private = password_verify($password, $hashprivate); 
     if ($hash_private != 0) { 

      $_SESSION['username'] = $rowprivate['username']; 
      $_SESSION['logged'] = true; 
      $_SESSION['business'] = $rowprivate['bname']; 
      $_SESSION['type'] = 'business'; 
     } else { 
      //no record 
     } 
    } 
?>

來源

2017-02-07 14:55:32 Akintunde007

一個數據庫的作品，第二不

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