将二进制转换为ASCII，ASCII中的旋转字符串-Java

Question

I'm really new in java and now I'm really lost. I have to first, convert the binary to its ascii. Then, create a rotation string (ex: "2L4R6L") of the ascii to get a specific letter.

I'm still on the first part but now I'm really lost. I tried the conversion, but when I print it, null is the output. Can you help me point out my mistake, and help me solve this program?

Here's the methods I created:

public void setEncryptedMessage(String encryptedMess){
    encryptedMessage = encryptedMess;
    Cipher cph = new Cipher();
    cph.convertBinary(encryptedMessage);
}

public void convertBinary(String encryptedMessage){
    StringTokenizer st = new StringTokenizer(encryptedMessage, '#');
    int convert = Integer.parseInt(st.nextToken(), 2);
    String letter = new Character((char)convert).toString();
    encryptedMessage = letter;
}   

public String getEncryptedMessage(){    
  return encryptedMessage;
}   

this is the main:

public static void main(String[] args){ 
  Cipher cph=new Cipher();
  String encryptedMessage="1000001#1001001#1011010#1010000#1000110";    
  cph.setEncryptedMessage(encryptedMessage);    
  System.out.println(cph.getEncryptedMessage());
}

Answer 1

摆脱在setEncryptedMessage创建的额外Cipher对象

Answer 2

So you are having trouble converting a Binary in string to ascii form, right?
Here! I found a fix!

public static void main(String[] args) {
    String encryptedMessage="1000001#1001001#1011010#1010000#1000110"; //BTW one ascii charecter is represented by 8 digits in binary. And here there are 7 digits per charecter...fix that and well moving on...
    String filtered= encryptedMessage.replaceAll("#", "");
    StringBuilder b = new StringBuilder();
    int i = 0;
    String rslt= "";
    while (i + 8 <= filtered.length()) {
    char c = convert(filtered.substring(i, i+8));
    i+=8;
    b.append(c);
    rslt= b.toString();
    }
    System.out.println(rslt);
}
private static char convert(String bs) {
    return (char)Integer.parseInt(bs, 2);
}