Tokenize字節數組

Question

我有一個原始字節數組，我需要將其標記爲java中的一個字節數組列表。以下方法聲明更好地解釋了這一點。

public static List<byte[]> splitMessage(byte[] rawByte, String tokenDelimiter) 

示例運行。

示例執行1：

原始字節

byte[] rawBytes = new byte[]{72,118,121,49,85,118,97,113,111,124,44,124,49,48,43,57,48,36,63,49,66,70,22,18,124,44,124,23,27,25,54,24,24,34,44,57,69,66,49,47,66,16,39,35,32,36,30,50,63,124,44,124,16,18,24,64,4,94,124,44,124,19,31,42,55,66,46,34,62,34,37}; 

tokenDelimiter是|,|（即124,44,124）

所以列表返回是：

Token 1: 72,118,121,49,85,118,97,113,111 
Token 2: 49,48,43,57,48,36,63,49,66,70,22,18 
Token 3: 23,27,25,54,24,24,34,44,57,69,66,49,47,66,16,39,35,32,36,30,50,63, 
Token 4: 16,18,24,64,4,94 
Token 5: 19,31,42,55,66,46,34,62,34,37 

示例執行2：

byte[] rawBytes = new byte[]{72,118,121,49,85,118,97,113,111,124,44,124,49,48,43,57,48,36,63,49,66,70,22,18,124,44,124,124,44,124,23,27,25,54,24,24,34,44,57,69,66,49,47,66,16,39,35,32,36,30,50,63,124,44,124,16,18,24,64,4,94,124,44,124,19,31,42,55,66,46,34,62,34,37,124,44,124,124,44,124}; 

tokenDelimiter是|,|（即124,44,124）

Token 1: 72,118,121,49,85,118,97,113,111 
Token 2: 49,48,43,57,48,36,63,49,66,70,22,18 
Token 3: <Empty> 
Token 3: 23,27,25,54,24,24,34,44,57,69,66,49,47,66,16,39,35,32,36,30,50,63, 
Token 4: 16,18,24,64,4,94 
Token 5: 19,31,42,55,66,46,34,62,34,37 
Token 6: <Empty> 
Token 7: <Empty> 

我能夠才達到例如從下面的代碼段中運行。但是在第二個標籤中卡住了。

public static List<byte[]> splitMessageSept19(byte[] rawByte, String tokenDelimiter) throws UnsupportedEncodingException 
{ 
    List<byte[]> tokens = new ArrayList<byte[]>(); 

    final byte[] byteArray = tokenDelimiter.getBytes("UTF-8"); 
    final byte byteDelimitorFirstByte = byteArray[0]; 

    int bytenum =0 ; 
    int lastIndex = 0; 
    int storIterator =0; 
    for (int iterator = 0 ; iterator <= rawByte.length ; iterator++) 
    { 
     if (iterator == rawByte.length || rawByte[iterator] == byteDelimitorFirstByte) 
     { 
      storIterator = iterator; 
      if (iterator != rawByte.length) 
      { 
       for (int i=0 ; i < byteArray.length ; i++) 
       { 
        if (rawByte[iterator] == byteArray[i]) 
        { 
         iterator++ ; 
         continue; 
        } 
        else 
        { 
         break; 
        } 
       } 
      } 
      byte[] byteArrayExtracted = new byte[storIterator - lastIndex]; 
      System.arraycopy(rawByte, lastIndex, byteArrayExtracted, 0, 
          storIterator - lastIndex); 
      lastIndex = iterator ; 
      tokens.add(byteArrayExtracted); 
      byteArrayExtracted = null; 
     } 
    } 
    for (byte[] bytetoken : tokens) 
    { 
     System.out.println("Token received is: " + new String(bytetoken, "UTF-8")); 
    } 
    return tokens; 
} 

有沒有人遇到類似的令牌化數組問題？請建議是否有其他方法來標記數組。

請注意：我不想將字節流轉換爲String，以字符串格式進行標記並轉換回字節。它可能存在編碼問題。

Answer 1

如果您使用ISO-8859-1，那麼字節會保留原樣。

private static final Charset ISO_8859_1 = Charset.forName("ISO-8859-1"); 

public static List<byte[]> splitMessageSept19(byte[] rawByte, String tokenDelimiter) { 
    Pattern pattern = Pattern.compile(tokenDelimiter, Pattern.LITERAL); 
    String[] parts = pattern.split(new String(rawByte, ISO_8859_1), -1); 
    List<byte[]> ret = new ArrayList<byte[]>(); 
    for (String part : parts) 
     ret.add(part.getBytes(ISO_8859_1)); 
    return ret; 
} 

public static void main(String... args) { 
    StringBuilder sb = new StringBuilder(); 
    for(int i=0;i<256;i++) 
     sb.append((char) i); 
    byte[] bytes = sb.toString().getBytes(ISO_8859_1); 
    List<byte[]> list = splitMessageSept19(bytes, ","); 
    for (byte[] b : list) 
     System.out.println(Arrays.toString(b)); 
} 

打印

[0，1，2，3，4，5，6，7，8，9，10，11，12，13，14，15，16，17，18 ，19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43號決議] [45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68 ，69,70,71,72,73,74,75,66,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93 ，94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118 ，119,120,121,122,123,124,125,126,127，-128，-127，-126，-125，-124，-123，-122，-121，-120，-119， - 118，-117，-116，-115，-114，-113，-112， -111，-110，-109，-108，-107，-106，-105，-104，-103，-102，-101，-100，-99，-98，-97，-96，-95 ，-94，-93，-92，-91，-90，-89，-88，-87，-86，-85，-84，-83，-82，-81，-80，-79， - -78，-77，-76，-75，-74，-73，-72，-71，-70，-69，-68，-67，-66，-65，-64，-63，-62， -61，-60，-59，-58，-57，-56，-55，-54，-53，-52，-51，-50，-49，-48，-47，-46，-45 ，-44，-43，-42，-41，-40，-39，-38，-37，-36，-35，-34，-33，-32，-31，-30，-29， - 28，-27，-26，-25，-24，-23，-22，-21，-20，-19，-18，-17，-16，-15，-14，-13，-12， -11，-10，-9，-8，-7，-6，-5，-4，-3，-2，-1]

調用

byte[] rawBytes = new byte[]{72,118,121,49,85,118,97,113,111,124,44,124,49,48,43,57,48,36,63,49,66,70,22,18,124,44,124,124,44,124,23,27,25,54,24,24,34,44,57,69,66,49,47,66,16,39,35,32,36,30,50,63,124,44,124,16,18,24,64,4,94,124,44,124,19,31,42,55,66,46,34,62,34,37,124,44,124,124,44,124}; 
List<byte[]> list = splitMessageSept19(rawBytes, "|,|"); 

產生

[72, 118, 121, 49, 85, 118, 97, 113, 111] 
[49, 48, 43, 57, 48, 36, 63, 49, 66, 70, 22, 18] 
[] 
[23, 27, 25, 54, 24, 24, 34, 44, 57, 69, 66, 49, 47, 66, 16, 39, 35, 32, 36, 30, 50, 63] 
[16, 18, 24, 64, 4, 94] 
[19, 31, 42, 55, 66, 46, 34, 62, 34, 37] 
[] 
[] 

Answer 2

您應該看看KMP算法：KMP on wikipedia以及其他字符串搜索算法。

作爲一個快速解決您的代碼試試這個：

public static List<byte[]> splitMessageSept19(byte[] rawByte, String tokenDelimiter) throws UnsupportedEncodingException 
{ 
    List<byte[]> tokens = new ArrayList<byte[]>(); 

    final byte[] byteArray = tokenDelimiter.getBytes("UTF-8"); 
    int lastIndex = 0; 

    for (int iterator = 0; iterator < rawByte.length - byteArray.length + 1;) 
    { 
     boolean patternFound = true; 
     for (int i = 0; i < byteArray.length; i++) 
     { 
      if (rawByte[iterator + i] != byteArray[i]) 
      { 
       patternFound = false; 
       break; 
      } 
     } 
     if (patternFound) 
     { 
      byte[] byteArrayExtracted = new byte[iterator - lastIndex]; 
      System.arraycopy(rawByte, lastIndex, byteArrayExtracted, 0, iterator - lastIndex); 
      iterator += byteArray.length; 
      lastIndex = iterator; 
      tokens.add(byteArrayExtracted); 
     } 
     else 
      iterator++; 

    } 
    for (byte[] bytetoken : tokens) 
    { 
     System.out.println("Token received is: " + new String(bytetoken, "UTF-8")); 
    } 
    return tokens; 
} 

我還沒有編譯這個代碼，所以它可能被打破，但即使我希望你的想法。

這是一個天真的算法，它非常慢，特別是如果分隔符很長。如果你想要更好的東西，去看看其他一些字符串搜索算法。