PHP和Swift:密码_hash（）出现致命错误

我已经在服务器上设置了我的注册和登录php文件，它与我的swift应用程序配合使用。我可以很容易地登录和注册。但是，当我添加

password\u hash（）

方法对用户密码进行安全保护时，当我尝试注册时，它在Xcode上给出了一个错误。如果密码不再有效，是否有其他方法可以安全地存储密码。是的，我安装了php 5.5.34：

通过Xcode出错：

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 />

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;

}

您可以使用它来获取

密码*（）

函数。还提供您可以使用它来获取

密码*（）

函数。还提供

// 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;

}