Question

抱歉我的英文。我有方法解码字符串，但在控制台中我有warn。我得到代码字符串base64，我需要解码它的字符串。

06-11 06:49:40.557: W/EGL_genymotion(1611): eglSurfaceAttrib not implemented
06-11 06:49:41.041: W/System.err(1611): java.io.IOException: Bad Base64 input character decimal 208 in array position 0
06-11 06:49:41.041: W/System.err(1611):     at com.motottaxi24user.Base64.decode(Base64.java:1201)
06-11 06:49:41.077: W/System.err(1611):     at com.motottaxi24user.Base64.decode(Base64.java:1122)
06-11 06:49:41.077: W/System.err(1611):     at com.motottaxi24user.User.decodeString(User.java:240)
06-11 06:49:41.097: W/System.err(1611):     at com.motottaxi24user.User$NewsAsynkTask.doInBackground(User.java:200)
06-11 06:49:41.137: W/System.err(1611):     at com.motottaxi24user.User$NewsAsynkTask.doInBackground(User.java:1)
06-11 06:49:41.141: W/System.err(1611):     at android.os.AsyncTask$2.call(AsyncTask.java:287)
06-11 06:49:41.181: W/System.err(1611):     at java.util.concurrent.FutureTask.run(FutureTask.java:234)
06-11 06:49:41.197: W/System.err(1611):     at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
06-11 06:49:41.213: W/System.err(1611):     at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080)
06-11 06:49:41.217: W/System.err(1611):     at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573)
06-11 06:49:41.237: W/System.err(1611):     at java.lang.Thread.run(Thread.java:856)
06-11 06:49:41.281: W/System.err(1611): java.io.IOException: Bad Base64 input character decimal 58 in array position 2

我的方法：

public String decodeString(String line) {
        byte[] tmp2;
        String val2 = null;

        try {
            tmp2 = Base64.decode(line);
            val2 = new String(tmp2, "UTF-8");
        } catch (IOException e) {
            e.printStackTrace();
        }

        return val2;
    }

Answer 1

错误消息sais，在第二个位置的字符串中是代码为58的字符。这是冒号：在base64中不允许。所以，检查你的输入

Answer 2

尝试这样的事情：

public String decodeString(String line) {
  byte[] tmp2;
  String val2 = null;

  try {
      tmp2 = Base64.decode(line.getBytes());
      val2 = new String(Base64.encode(tmp2), "UTF-8");
  } catch (IOException e) {
      e.printStackTrace();
  }
  return val2;
}

或者看看Error when trying to encode/decode String to Base64

数组位置0中的Bad Base64输入字符十进制208（android）

2 个答案: