PHP： openssl_private_decrypt（）： 密钥参数不是有效的私钥

Question

我有这个代码来创建RSA 4096公钥和私钥来加密和解密字符串。法典：<?php$config = array(    "config" => "C:/xampp/php/extras/openssl/openssl.cnf",    "private_key_bits" => 4096,    "private_key_type" => OPENSSL_KEYTYPE_RSA);// Create the private and public key$res = openssl_pkey_new($config);// Extract the private key from $res to $privKeyopenssl_pkey_export($res, $privKey);// Extract the public key from $res to $pubKey$pubKey = openssl_pkey_get_details($res);$pubKey = $pubKey["key"];$data = 'Hello, World!';// Encrypt the data to $encrypted using the public keyopenssl_public_encrypt($data, $encrypted, $pubKey);echo $encrypted;// Decrypt the data using the private key and store the results in $decryptedopenssl_private_decrypt($encrypted, $decrypted, $privKey);echo $decrypted;?>它创建密钥，加密字符串 （），但当尝试解密字符串时，会发生错误：dataHello, World!encrypted警告：openssl_private_decrypt（）： 在第 26 行的 C：\xampp\htdocs\rsa\index.php 密钥参数不是有效的私钥

Answer 1

好吧，这对我有用：

更改为 。openssl_pkey_export($res, $privKey);openssl_pkey_export($res, $privKey, NULL, $config);

Answer 2

您不需要像这样导出密钥私有，至少在将其保存在安全的地方之前是这样：

$config = array(

    "private_key_bits" => 4096,

    "private_key_type" => OPENSSL_KEYTYPE_RSA

);

$pki     = openssl_pkey_new($config);

$public  = openssl_pkey_get_public(

    openssl_pkey_get_details($pki)['key']

); // why on earth did they implement it like this? so clunky.

$private = openssl_pkey_get_private($pki);

$data = 'Hello, World!';

openssl_public_encrypt($data, $encrypted, $public);

openssl_private_decrypt($encrypted, $decrypted, $private);

var_dump(

    bin2hex($encrypted),

    $decrypted

);