PHP和Swift：password_hash（）的致命错误

我已经设置了我的注册并登录了它们在服务器上的php文件，它适用于我的swift应用程序。我可以轻松登录并注册。但是当我添加password_hash()方法来保护用户密码时，当我尝试注册时，它会在Xcode上出现错误。有没有其他方法可以安全地存储密码，如果这不起作用了。是的，我的PHP安装34年5月5日：通过Xcode的PHP和Swift：password_hash（）的致命错误

错误：

DATA: <br /> 
<b>Fatal error</b>: Call to undefined function password_hash() in <b>/*/*/public_html/*/signup.php</b> on line <b>92</b><br />

signup.php

// Hash password and insert new user to table 

$hashPassword = password_hash($password, PASSWORD_DEFAULT); 
$command = " INSERT INTO USER 
       (firstname, lastname, username, email, password) 
      VALUES 
       ('$firstname', '$lastname', '$username', '$email', '$hashPassword')"; 

if (mysqli_query($DB, $command)) { 

    // Search for newUser 

    $command = "SELECT * FROM USER WHERE username = '$username'"; 
    $sql = mysqli_query($DB, $command); 

    if (mysqli_num_rows($sql) != 0) { 
     $newUser = mysqli_fetch_array($sql); 

     $returnData["status"]  = "200"; 
     $returnData["message"]  = "Success!"; 
     $returnData["ID"]   = $newUser["ID"]; 
     $returnData["firstname"] = $newUser["firstname"]; 
     $returnData["lastname"]   = $newUser["lastname"]; 
     $returnData["username"]   = $newUser["username"]; 
     $returnData["email"]  = $newUser["email"]; 

    } 

    echo json_encode($returnData); 
    return; 

} else { 

    $returnData["status"] = "400"; 
    $returnData["message"] = "Sorry, something must've went wrong. Please try again..."; 
    echo json_encode($returnData); 
    return; 

}

signin.php

// Find user from table and sign in 

$command = "SELECT * FROM USER WHERE email = '$email'"; 
$sql = mysqli_fetch_array(mysqli_query($DB, $command)); 

if (isset($sql)) { 

    $hashPassword = $sql["password"]; 

    if (password_verify($password, $hashPassword)) { 

     $returnData["status"] = "200"; 
     $returnData["message"] = "Success!"; 
     $returnData["ID"] = $sql["ID"]; 
     $returnData["username"] = $sql["username"]; 

    } 

    echo json_encode($returnData); 
    return;  

} else { 

    $returnData["status"] = "400"; 
    $returnData["message"] = "Sorry, something must've went wrong. Please try again..."; 

    echo json_encode($returnData); 
    return; 

}

来源

2016-06-12 Mohamed Mohamed

您可以使用此library来获取password_*()函数。还提供了< PHP 5.5的支持。

来源

2016-06-12 08:53:27 Nizarii

我也试过，它根本不起作用。 –

PHP和Swift：password_hash（）的致命错误

回答

相关问题