java无法在Android中解析为整数

Question

我想为一些String值实现encrypt和decrypt操作。我已正确加密，但我不明白如何decrypt此值如下：

jsonString Values ={"Response":"NJGOkF2EvOIpfKG14LHQZrVfj\/OEJvopi+OKU+q5G2ynDbVUnIckfMLGCCsxcY9+BmVg+KJXF1ls\nGf2rWg73iyowyq6THyDfBS8uZnSp9PfS3bJCFb6YWX4\/\/uxjDwtZ","statusFlag":"true"}

当我解密时，我得到NumberFormatException

这是我的简单加密类

import java.security.SecureRandom;
import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.SecretKey;
import javax.crypto.spec.SecretKeySpec;


    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(); //2 * i, 2 * i + 2
                    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));
            }

    }

这是日志猫信息

01-03 11:30:51.154: W/System.err(437): java.lang.NumberFormatException: unable to parse '{"' as integer
01-03 11:30:51.164: W/System.err(437):  at java.lang.Integer.parse(Integer.java:383)
01-03 11:30:51.164: W/System.err(437):  at java.lang.Integer.parseInt(Integer.java:372)
01-03 11:30:51.164: W/System.err(437):  at java.lang.Integer.valueOf(Integer.java:528)
01-03 11:30:51.164: W/System.err(437):  at com.json_to_server.SimpleCrypto.toByte(SimpleCrypto.java:63)
01-03 11:30:51.164: W/System.err(437):  at com.json_to_server.SimpleCrypto.decrypt(SimpleCrypto.java:20)
01-03 11:30:51.164: W/System.err(437):  at com.json_to_server.EncryptDecrypt_Demo.POST(EncryptDecrypt_Demo.java:202)
01-03 11:30:51.174: W/System.err(437):  at com.json_to_server.EncryptDecrypt_Demo$HttpAsyncTask.doInBackground(EncryptDecrypt_Demo.java:267)
01-03 11:30:51.174: W/System.err(437):  at com.json_to_server.EncryptDecrypt_Demo$HttpAsyncTask.doInBackground(EncryptDecrypt_Demo.java:1)
01-03 11:30:51.174: W/System.err(437):  at 安卓.os.AsyncTask$2.call(AsyncTask.java:185)
01-03 11:30:51.174: W/System.err(437):  at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:306)
01-03 11:30:51.174: W/System.err(437):  at java.util.concurrent.FutureTask.run(FutureTask.java:138)
01-03 11:30:51.174: W/System.err(437):  at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1088)
01-03 11:30:51.174: W/System.err(437):  at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:581)
01-03 11:30:51.174: W/System.err(437):  at java.lang.Thread.run(Thread.java:1019)

我想解密这个值。我不知道我必须在哪里更改我的SimpleCrypto class

jsonString Values = {"Response":"NJGOkF2EvOIpfKG14LHQZrVfj\/OEJvopi+OKU+q5G2ynDbVUnIckfMLGCCsxcY9+BmVg+KJXF1ls\nGf2rWg73iyowyq6THyDfBS8uZnSp9PfS3bJCFb6YWX4\/\/uxjDwtZ","statusFlag":"true"}

Answer 1

我假设你是decrypting整个JSON字符串，我想你只需要decrypt

此字符串： NJGOkF2EvOIpfKG14LHQZrVfj\/OEJvopi+OKU+q5G2ynDbVUnIckfMLGCCsxcY9+BmVg+KJXF1ls\nGf2rWg73iyowyq6THyDfBS8uZnSp9PfS3bJCFb6YWX4\/\/uxjDwtZ

所以你应该parse从JSON得到Response string，然后decrypt它

更新： 您可以像这样解析json字符串：

String value = "YOUR_JSON_STRING";
    try {
        JSONObject mJsonObject = new JSONObject(value);
        String response = mJsonObject.getString("Response");
        //Decrypt response string 

    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

现在你的encryptedstring在response variable中。现在你可以decrypt{}

Answer 2

您正在尝试将整数字符串“NJ”解析为十六进制数字。但是十六进制只有0-9和A-F符号。将您的收入字符串更改为仅十六进制符号

Answer 3

您正在使用JSON数据作为字符串数据。 您只需要选择JSON的“响应”部分值，如下所示

JSONObject jO= new JSONObject(value);
String response = jO.getString("Response");

Answer 4

我认为result[i] = Integer.valueOf(hexString.substring(2*i, 2*i+2), 16).byteValue();会给您带来错误，因为您的十六进制字符串可能包含字符literal so Integer。valueOf（）函数提供NumberFormatException

有关hexstring到字节数组的转换，请检查[link]https://stackoverflow.com/a/140861/3131537

希望这能解决你的问题