无法解密在Android应用程序中的字符串 [英] Unable to decrypt String in android App
问题描述
我试图开发一个可以加密和解密值的Android应用程序。因此，我已关注此链接在此输入链接描述
I was trying to develop an android application that could encrypt and decrypt values. So I have followed this link enter link description here
到目前为止，我能够加密文本，但我无法解密。在我的代码中，我使用了提供的链接中提到的相同的AESHelper类。
So far I was able to encrypt a text but I was not able to decrypt it. In my code I have used the same AESHelper class which is mentioned in the link provided.
下面是我的活动类，我已经用于加密和解密值
The below is my activity class that i have used to Encrypt and decrypt the values
import android.support.v7.app.ActionBarActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;
public class MainActivity extends ActionBarActivity {
EditText text ;
TextView encp,decriptom;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
text = (EditText) findViewById(R.id.editText);
encp = (TextView) findViewById(R.id.valueexcript);
decriptom = (TextView) findViewById(R.id.deexcript);
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
public void Ecript(View v)
{
String Key = "avc";
try {
String normalTextEnc = AHShelper.encrypt(Key, text.getText().toString());
Toast.makeText(this,normalTextEnc,Toast.LENGTH_LONG).show();
encp.setText(normalTextEnc);
} catch (Exception e) {
e.printStackTrace();
}
// Toast.makeText(this,"Hello",Toast.LENGTH_LONG).show();
String decript;
try {
decript = AHShelper.decrypt(Key,encp.getText().toString());
decriptom.setText(decript);
Toast.makeText(this,decript,Toast.LENGTH_LONG).show();
} catch (Exception e) {
e.printStackTrace();
}
}
}
我使用的AHShelper类如下
AHShelper class that i have used is below
import java.security.SecureRandom;
import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.SecretKey;
import javax.crypto.spec.SecretKeySpec;
public class AHShelper {
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));
}
}
推荐答案
基本上，这个代码依赖一个小技巧：如果在使用之前为SUN提供程序和Bouncy Castle提供程序种子SHA1PRNG，则它将始终生成相同的随机字节流。这并不总是每个提供者的情况;其他提供商只是混合在种子。换句话说，它们可以使用预接种的PRNG。在这种情况下， getRawKey 方法会为加密和解密生成不同的密钥，这将导致解密失败。
Basically this code relies on a little trick: if you seed the SHA1PRNG for the SUN provider and Bouncy Castle provider before it is used then it will always generate the same stream of random bytes. This is not always the case for every provider though; other providers simply mix in the seed. In other words, they may use a pre-seeded PRNG. In that case the getRawKey method generates different keys for the encrypt and decrypt, which will result in a failure to decrypt.
基本上，这个可怕的代码片段滥用SHA1PRNG作为密钥导出Fucntion或KDF。如果输入是密码，则应使用真正的KDF，例如PBKDF2;如果输入是密钥，则应使用HKDF。
Basically this horrible code snippet abuses the SHA1PRNG as a Key Derivation Fucntion or KDF. You should use a true KDF such as PBKDF2 if the input is a password or HKDF if the input is a key.
应删除该代码片段。它已从Android代码段复制过来，但我找不到该网站了。
That code snippet should be removed. It has been copied from Android snippets, but I cannot find that site anymore.
这篇关于无法解密在Android应用程序中的字符串的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
