为了减少对DB的调用次数,我尝试将此代码修改为java:
--set serveroutput on
declare
l_aux NUMBER;
l_cle RAW(9) := utl_raw.cast_to_raw('example21');
l_crypt_raw VARCHAR2(200);
l_crypt_str VARCHAR2(200);
p_txt_desencrip varchar2(200):='8387F8937F5F842F805C44B88429D2CD';
BEGIN
l_crypt_raw := utl_raw.cast_to_raw(utl_raw.cast_to_varchar2( p_txt_desencrip));
DBMS_OBFUSCATION_TOOLKIT.DESDECRYPT ( input => p_txt_desencrip
, key => l_cle
, decrypted_data => l_crypt_raw
);
l_crypt_str := utl_raw.cast_to_varchar2(l_crypt_raw);
l_aux := LENGTH(l_crypt_str);
l_crypt_str := RPAD(l_crypt_str,l_aux-ASCII(SUBSTR(l_crypt_str,l_aux)));
DBMS_OUTPUT.PUT_LINE('Decypted message->' || l_crypt_str);
END;我已经看到了所有解决这一任务的脚本,但无论如何,我的主要问题是,我的钥匙有9个字符。以下是我的java代码:
import java.security.InvalidKeyException;
import java.security.NoSuchAlgorithmException;
import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.NoSuchPaddingException;
import javax.crypto.SecretKey;
import javax.crypto.spec.SecretKeySpec;
import com.sun.mail.util.BASE64DecoderStream;
import com.sun.mail.util.BASE64EncoderStream;
public class Test{
private static Cipher ecipher;
private static Cipher dcipher;
private static SecretKey key;
public static void main(String[] args) {
try {
String clave = "example21";
// generate secret key using DES algorithm
SecretKey key2 = new SecretKeySpec(clave.getBytes(), 0, 9, "DES");
ecipher = Cipher.getInstance("DES");
dcipher = Cipher.getInstance("DES");
// initialize the ciphers with the given key
ecipher.init(Cipher.ENCRYPT_MODE, key2);
dcipher.init(Cipher.DECRYPT_MODE, key2);
String encrypted = encrypt("text to encrypt");
System.out.println(encrypted);
String decrypted = decrypt(encrypted);
System.out.println("Decrypted: " + decrypted);
}catch (NoSuchAlgorithmException e) {
System.out.println("No Such Algorithm:" + e.getMessage());
return;
}
catch (NoSuchPaddingException e) {
System.out.println("No Such Padding:" + e.getMessage());
return;
}
catch (InvalidKeyException e) {
System.out.println("Invalid Key:" + e.getMessage());
return;
}
}
public static String encrypt(String str) {
try {
// encode the string into a sequence of bytes using the named charset
// storing the result into a new byte array.
byte[] utf8 = str.getBytes("UTF8");
byte[] enc = ecipher.doFinal(utf8);
// encode to base64
enc = BASE64EncoderStream.encode(enc);
return new String(enc);
}catch (Exception e) {
e.printStackTrace();
}
return null;
}
public static String decrypt(String str) {
try {
// decode with base64 to get bytes
byte[] dec = BASE64DecoderStream.decode(str.getBytes());
byte[] utf8 = dcipher.doFinal(dec);
// create new string based on the specified charset
return new String(utf8, "UTF8");
} catch (Exception e) {
e.printStackTrace();
}
return null;
}
}我尝试过不同的方法和算法,但我总是得到错误的“无效密钥:错误的密钥大小”.
有什么建议我应该试试吗?
提前感谢!
发布于 2021-06-21 18:05:34
DBMS_OBFUSCATION_TOOLKIT被废弃,不应该是used.DES,它在20多年前就被破坏了,不应该再被使用了。至于你的问题。关于关键参数,DBMS_OBFUSCATION_TOOLKIT.DESDECRYPT的文档有以下内容:
如果键长度丢失或小于8个字节,则该过程将引发错误ORA-28234“键长度太短”。注意,如果使用较大的键,则会忽略额外的字节。,这样一个9字节的键就不会生成异常.
因此,Java只是对密钥进行了更多的验证。
https://stackoverflow.com/questions/68065933
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