6

我有一个原始字节数组，我需要将其标记为 java 中的字节数组列表。通过以下方法声明更好地解释。

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

示例运行。

示例运行 1：

原始字节

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）

所以返回的列表如下：

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

示例运行 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>

我能够从以下代码片段中运行示例。但在第二个中卡住了标签。

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;
}

有没有人遇到过类似的标记数组的问题？请建议是否有其他方法来标记数组。

请注意：我不想将字节流转换为String，以字符串格式标记化并转换回字节。它可能有编码问题。

4

2 回答 2

3

如果您使用 ISO-8859-1，则字节会按原样保留。

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));
}

印刷

[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] [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, 76, 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]

打电话

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, "|,|");

生产

[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]
[]
[]
于 2012-09-19T10:28:52.237 回答
0

您应该查看 KMP 算法：维基百科上的 KMP和其他字符串搜索算法。

作为对您的代码的快速修复，请尝试以下操作：

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;
}

我还没有编译这段代码，所以它可能会被破坏，但即使我希望你明白这一点。

这是一个很简单的算法，特别是在分隔符很长的情况下。如果您想要更好的东西，请查看其他一些字符串搜索算法。

于 2012-09-19T10:46:59.683 回答