PHP和Swift : password_hash()上的致命错误

Question

我已经注册并登录了php文件，它们在服务器上，它可以与我的swift应用程序一起工作。我可以轻松地登录和注册。但是，当我将password_hash()方法添加到用户密码的had安全性时，当我尝试注册时，它在Xcode上给出一个错误。如果密码不再起作用，有没有其他方法可以安全地存储密码。是的，我已经安装了php 5.5.34：

通过Xcode产生的错误：

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;

}

Answer 1

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