【发布时间】：2011-09-23 02:07:05
【问题描述】：
我使用加密 AES 算法，当我加密 16 字节（一个块）时，结果是 32 字节。 这样可以吗？
我使用的源代码是：
package net.sf.andhsli.hotspotlogin;
import java.security.SecureRandom;
import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.SecretKey;
import javax.crypto.spec.SecretKeySpec;
/**
* Usage:
* <pre>
* String crypto = SimpleCrypto.encrypt(masterpassword, cleartext)
* ...
* String cleartext = SimpleCrypto.decrypt(masterpassword, crypto)
* </pre>
* @author ferenc.hechler
*/
public class SimpleCrypto {
public static String encrypt(String seed, String cleartext) throws Exception {
byte[] rawKey = getRawKey(seed.getBytes());
byte[] result = encrypt(rawKey, cleartext.getBytes());
return toHex(result);
}
public static String decrypt(String seed, String encrypted) throws Exception {
byte[] rawKey = getRawKey(seed.getBytes());
byte[] enc = toByte(encrypted);
byte[] result = decrypt(rawKey, enc);
return new String(result);
}
private static byte[] getRawKey(byte[] seed) throws Exception {
KeyGenerator kgen = KeyGenerator.getInstance("AES");
SecureRandom sr = SecureRandom.getInstance("SHA1PRNG");
sr.setSeed(seed);
kgen.init(128, sr); // 192 and 256 bits may not be available
SecretKey skey = kgen.generateKey();
byte[] raw = skey.getEncoded();
return raw;
}
private static byte[] encrypt(byte[] raw, byte[] clear) throws Exception {
SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
Cipher cipher = Cipher.getInstance("AES");
cipher.init(Cipher.ENCRYPT_MODE, skeySpec);
byte[] encrypted = cipher.doFinal(clear);
return encrypted;
}
private static byte[] decrypt(byte[] raw, byte[] encrypted) throws Exception {
SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
Cipher cipher = Cipher.getInstance("AES");
cipher.init(Cipher.DECRYPT_MODE, skeySpec);
byte[] decrypted = cipher.doFinal(encrypted);
return decrypted;
}
public static String toHex(String txt) {
return toHex(txt.getBytes());
}
public static String fromHex(String hex) {
return new String(toByte(hex));
}
public static byte[] toByte(String hexString) {
int len = hexString.length()/2;
byte[] result = new byte[len];
for (int i = 0; i < len; i++)
result[i] = Integer.valueOf(hexString.substring(2*i, 2*i+2), 16).byteValue();
return result;
}
public static String toHex(byte[] buf) {
if (buf == null)
return "";
StringBuffer result = new StringBuffer(2*buf.length);
for (int i = 0; i < buf.length; i++) {
appendHex(result, buf[i]);
}
return result.toString();
}
private final static String HEX = "0123456789ABCDEF";
private static void appendHex(StringBuffer sb, byte b) {
sb.append(HEX.charAt((b>>4)&0x0f)).append(HEX.charAt(b&0x0f));
}
}
【问题讨论】：
-
我删除了 Android 标签，因为这与它没有任何关系。顺便提一句。你真的应该给我们一个例子，说明你如何使用这个类来说明问题所在。
-
这真的不属于 codereview.SE 吗？
-
你得到 32 个字符还是 64 个字符？
-
我假设您使用的是十六进制编码，因为您想要 URI 友好的字符。在这种情况下，我建议使用 base32 编码。
-
这将不再适用于 Android 4.2，这里有很好的信息：stackoverflow.com/questions/13433529/…
标签： java encryption aes