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 $privKey

openssl_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 key

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

echo $encrypted;

// Decrypt the data using the private key and store the results in $decrypted

openssl_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：$config = array(&nbsp; &nbsp; "private_key_bits" => 4096,&nbsp; &nbsp; "private_key_type" => OPENSSL_KEYTYPE_RSA);$pki&nbsp; &nbsp; &nbsp;= openssl_pkey_new($config);$public&nbsp; = openssl_pkey_get_public(&nbsp; &nbsp; 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, $private);var_dump(&nbsp; &nbsp; bin2hex($encrypted),&nbsp; &nbsp; $decrypted);