Question

已经给了我Java的示例代码，我不得不用PHP编写代码。 PHP代码的输出与给定的Java示例输出不匹配。示例Java代码可能并不完整（这是提供给我们的API文档中提供的内容。但是本质上，它总结了我们需要完成的步骤）

要加密的消息是 4NPMCXGERGXSBI7F9IPM5000004NPMCXGERGXSBI7F9IPM

JAVA代码

import javax.crypto.Cipher;</br>
import javax.crypto.spec.SecretKeySpec;</br>
import javax.crypto.spec.IvParameterSpec;</br>
import io.fabric.sdk.android.services.network.HttpRequest;</br>
class AAAA{  // class can have any name</br>
public static String getEncryptedTextFor(String text)</br>
    throws Exception {</br>
    Cipher cipher = Cipher.getInstance("AES/CBC/PKCS5Padding");</br>
    byte[] keyBytes = new byte[16];</br>
    String key = "YASHSHAH!@#987MOBAPP";</br>
    byte[] b = key.getBytes("UTF-8");</br>
    int len = b.length;</br>
    if (len > keyBytes.length) len = keyBytes.length;</br>
    System.arraycopy(b, 0, keyBytes, 0, len);</br>
    SecretKeySpec keySpec = new SecretKeySpec(keyBytes, "AES");</br>
    IvParameterSpec ivSpec = new IvParameterSpec(keyBytes);</br>
    cipher.init(Cipher.ENCRYPT_MODE, keySpec, ivSpec);</br>
    byte[] results = cipher.doFinal(text.getBytes("UTF-8"));</br>
    return HttpRequest.Base64.encodeBytes(results);</br>
    }</br>
    }</br>

我们尝试过的PHP代码

    $keyBytes = [];  
    $reqArrLen = 16;  
    $secretKey = 'YASHSHAH!@#987MOBAPP';  
    $a = utf8_encode($secretKey);  
    $b = unpack('C*', $a);   
    $len = count($b);  
    if($len > $reqArrLen){
    $len = $reqArrLen;
    }
    foreach ($b as $key => $value) {
    if($key <= $len)
    $keyBytes[]=$value;
    }
    echo '<pre>';
    $strtobyt = implode(array_map("chr", $keyBytes));
    $enLoan_Amount = AesCipher::encrypt($strtobyt, $eLoan_Amount);

Java中的示例输出格式为：（这不是实际输出，但是格式是共享的） pnehvlPu2gKJXJabR + HbMERdHWnrdc713 / SAsHOFWDQ9kv / eWNFI4IRmYWQ6QcoH

I'm not able to get this output format after executing PHP code.

示例PHP输出：

2438bf468c0142c82127d1dd740f53c3c19125f724265d58811322b50027a876a05b35f61a767fced426597d09f592f3

PHP输出长度为128，而Java输出长度为64。

使用AES / CBC / PKCS5Padding的值加密（Java代码转换为PHP）

0 个答案: