在 Java 中，如何将字节数组转换为十六进制数字字符串，同时保留前导零？

Question

我正在使用一些示例 java 代码来生成 md5 哈希值。 其中一部分将结果从字节转换为十六进制数字字符串：

byte messageDigest[] = algorithm.digest();     
StringBuffer hexString = new StringBuffer();
for (int i=0;i<messageDigest.length;i++) {
    hexString.append(Integer.toHexString(0xFF & messageDigest[i]));
    }

但是，它不太有效，因为 toHexString 显然会丢失前导零。 那么，从字节数组转换为保留前导零的十六进制字符串的最简单方法是什么？

Answer 1

这是一个错误的解决方案吗？ (android java)

    // Create MD5 Hash
    MessageDigest digest = java.security.MessageDigest.getInstance("MD5");
    digest.update(s.getBytes());
    byte[] md5sum = digest.digest();
    BigInteger bigInt = new BigInteger(1, md5sum);
    String stringMD5 = bigInt.toString(16);
    // Fill to 32 chars
    stringMD5 = String.format("%32s", stringMD5).replace(' ', '0');
    return stringMD5;

所以基本上它用 0 替换空格。

Answer 2

令我惊讶的是，没有人想出以下解决方案：

StringWriter sw = new StringWriter();
com.sun.corba.se.impl.orbutil.HexOutputStream hex = new com.sun.corba.se.impl.orbutil.HexOutputStream(sw);
hex.write(byteArray);
System.out.println(sw.toString());

Answer 3

或者你可以这样做：

byte[] digest = algorithm.digest();
StringBuilder byteContet = new StringBuilder();
for(byte b: digest){
 byteContent = String.format("%02x",b);
 byteContent.append(byteContent);
}

它简短、简单，基本上只是格式的改变。

Answer 4

这也是等效的，但使用 Apache util HexBin 其中代码简化为

HexBin.encode(messageDigest).toLowerCase();

Answer 5

查看 Apache Commons 编解码器的 ">Hex.encodeHexString。

import org.apache.commons.codec.binary.Hex;

String hex = Hex.encodeHexString(bytes);

Answer 6

您可以使用下面的一个。 我使用前导零字节和初始负字节对此进行了测试。

public static String toHex(byte[] bytes) {
    BigInteger bi = new BigInteger(1, bytes);
    return String.format("%0" + (bytes.length << 1) + "X", bi);
}

如果您想要小写十六进制数字，请在格式字符串中使用 "x" 。

Answer 7

一种简单的方法是检查 Integer.toHexString() 输出了多少位数字，并在需要时向每个字节添加前导零。 像这样的事情：

public static String toHexString(byte[] bytes) {
    StringBuilder hexString = new StringBuilder();

    for (int i = 0; i < bytes.length; i++) {
        String hex = Integer.toHexString(0xFF & bytes[i]);
        if (hex.length() == 1) {
            hexString.append('0');
        }
        hexString.append(hex);
    }

    return hexString.toString();
}

Answer 8

该方法 < strong>javax.xml.bind.DatatypeConverter.printHexBinary()，Java XML 绑定架构 (JAXB) 是将 byte[] 转换为十六进制字符串的便捷方法。 DatatypeConverter< /a> 类还包含许多其他有用的数据操作方法。

在 Java 8 及更早版本中，JAXB 是 Java 标准库的一部分。 它是 ，Java 11 中已删除作为将所有 Java EE 包移至其自己的库中的努力的一部分。 说来话长< /a>. 现在，javax.xml.bind 不存在，如果您想使用包含 DatatypeConverter 的 JAXB，则需要安装 JAXB API 和 JAXB 运行时。

用法示例：

byte bytes[] = {(byte)0, (byte)0, (byte)134, (byte)0, (byte)61};
String hex = javax.xml.bind.DatatypeConverter.printHexBinary(bytes);

将导致：

000086003D

Answer 9

我喜欢史蒂夫提交的内容，但他本来可以在没有几个变量的情况下完成，并在此过程中节省了几行。

public static String toHexString(byte[] bytes) {
    char[] hexArray = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    char[] hexChars = new char[bytes.length * 2];
    int v;
    for ( int j = 0; j < bytes.length; j++ ) {
        v = bytes[j] & 0xFF;
        hexChars[j*2] = hexArray[v/16];
        hexChars[j*2 + 1] = hexArray[v%16];
    }
    return new String(hexChars);
}

我喜欢这一点的是，很容易准确地看到它在做什么（而不是依赖于一些神奇的 BigInteger 黑盒转换），而且您也不必担心像前导零之类的极端情况。 该例程获取每个 4 位半字节并将其转换为十六进制字符。 它使用表查找，所以速度可能很快。 如果用按位移位和 AND 替换 v/16 和 v%16 可能会更快，但我现在懒得测试它。

Answer 10

我发现 Integer.toHexString 有点慢。 如果要转换许多字节，您可能需要考虑构建一个包含“00”..“FF”的字符串数组，并使用整数作为索引。 即，

hexString.append(hexArray[0xFF & messageDigest[i]]);

这更快并确保正确的长度。 只需要字符串数组：

String[] hexArray = {
"00","01","02","03","04","05","06","07","08","09","0A","0B","0C","0D","0E","0F",
"10","11","12","13","14","15","16","17","18","19","1A","1B","1C","1D","1E","1F",
"20","21","22","23","24","25","26","27","28","29","2A","2B","2C","2D","2E","2F",
"30","31","32","33","34","35","36","37","38","39","3A","3B","3C","3D","3E","3F",
"40","41","42","43","44","45","46","47","48","49","4A","4B","4C","4D","4E","4F",
"50","51","52","53","54","55","56","57","58","59","5A","5B","5C","5D","5E","5F",
"60","61","62","63","64","65","66","67","68","69","6A","6B","6C","6D","6E","6F",
"70","71","72","73","74","75","76","77","78","79","7A","7B","7C","7D","7E","7F",
"80","81","82","83","84","85","86","87","88","89","8A","8B","8C","8D","8E","8F",
"90","91","92","93","94","95","96","97","98","99","9A","9B","9C","9D","9E","9F",
"A0","A1","A2","A3","A4","A5","A6","A7","A8","A9","AA","AB","AC","AD","AE","AF",
"B0","B1","B2","B3","B4","B5","B6","B7","B8","B9","BA","BB","BC","BD","BE","BF",
"C0","C1","C2","C3","C4","C5","C6","C7","C8","C9","CA","CB","CC","CD","CE","CF",
"D0","D1","D2","D3","D4","D5","D6","D7","D8","D9","DA","DB","DC","DD","DE","DF",
"E0","E1","E2","E3","E4","E5","E6","E7","E8","E9","EA","EB","EC","ED","EE","EF",
"F0","F1","F2","F3","F4","F5","F6","F7","F8","F9","FA","FB","FC","FD","FE","FF"};

Answer 11

我一直在寻找同样的事情......这里有一些好主意，但我运行了一些微观基准测试。 我发现以下是最快的（从 Ayman 的上面修改而来，速度大约是 2 倍，比 Steve 的速度快 50%）：

public static String hash(String text, String algorithm)
        throws NoSuchAlgorithmException {
    byte[] hash = MessageDigest.getInstance(algorithm).digest(text.getBytes());
    return new BigInteger(1, hash).toString(16);
}

编辑：哎呀 - 错过了这本质上与 kgiannakakis 的相同，因此可能会被剥离前导 0。不过，将其修改为以下内容，它仍然是最快的：

public static String hash(String text, String algorithm)
        throws NoSuchAlgorithmException {
    byte[] hash = MessageDigest.getInstance(algorithm).digest(text.getBytes());
    BigInteger bi = new BigInteger(1, hash);
    String result = bi.toString(16);
    if (result.length() % 2 != 0) {
        return "0" + result;
    }
    return result;
}

Answer 12

我会使用类似这样的固定长度，例如散列：

md5sum = String.format("%032x", new BigInteger(1, md.digest()));

掩码中的 0 进行填充...

Answer 13

static String toHex(byte[] digest) {
    StringBuilder sb = new StringBuilder();
    for (byte b : digest) {
        sb.append(String.format("%1$02X", b));
    }

    return sb.toString();
}

Answer 14

String result = String.format("%0" + messageDigest.length + "s", hexString.toString())

鉴于您已有的解决方案，这是最短的解决方案。 如果您可以将字节数组转换为数值，String.format 可以同时将其转换为十六进制字符串。

Answer 15

Guava 也让它变得非常简单：

BaseEncoding.base16().encode( bytes );

当 Apache Commons 运行时，这是一个不错的选择无法使用。 它还有一些很好的输出控制，例如：

byte[] bytes = new byte[] { 0xa, 0xb, 0xc, 0xd, 0xe, 0xf };
BaseEncoding.base16().lowerCase().withSeparator( ":", 2 ).encode( bytes );
// "0a:0b:0c:0d:0e:0f"

Answer 16

这个解决方案有点老派，并且应该具有内存效率。

public static String toHexString(byte bytes[]) {
    if (bytes == null) {
        return null;
    }

    StringBuffer sb = new StringBuffer();
    for (int iter = 0; iter < bytes.length; iter++) {
        byte high = (byte) ( (bytes[iter] & 0xf0) >> 4);
        byte low =  (byte)   (bytes[iter] & 0x0f);
        sb.append(nibble2char(high));
        sb.append(nibble2char(low));
    }

    return sb.toString();
}

private static char nibble2char(byte b) {
    byte nibble = (byte) (b & 0x0f);
    if (nibble < 10) {
        return (char) ('0' + nibble);
    }
    return (char) ('a' + nibble - 10);
}

Answer 17

另外一个选择

public static String toHexString(byte[]bytes) {
    StringBuilder sb = new StringBuilder(bytes.length*2);
    for(byte b: bytes)
      sb.append(Integer.toHexString(b+0x800).substring(1));
    return sb.toString();
}

Answer 18

为了保持前导零，这里是 Paul 建议的一个小变化（例如 md5 哈希）：

public static String MD5hash(String text) throws NoSuchAlgorithmException {
    byte[] hash = MessageDigest.getInstance("MD5").digest(text.getBytes());
    return String.format("%032x",new BigInteger(1, hash));
}

哎呀，这看起来比 Ayman 提出的要差，对此感到抱歉

Answer 19

static String toHex(byte[] digest) {
    String digits = "0123456789abcdef";
    StringBuilder sb = new StringBuilder(digest.length * 2);
    for (byte b : digest) {
        int bi = b & 0xff;
        sb.append(digits.charAt(bi >> 4));
        sb.append(digits.charAt(bi & 0xf));
    }
    return sb.toString();
}

Answer 20

看起来 concat 和append 函数可能非常慢。 以下对我来说要快得多（比我之前的帖子）。 在构建输出时更改为字符数组是加快速度的关键因素。 我没有与 Brandon DuRette 建议的 Hex.encodeHex 进行比较。

public static String toHexString(byte[] bytes) {
    char[] hexArray = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    char[] hexChars = new char[10000000];
    int c = 0;
    int v;
    for ( j = 0; j < bytes.length; j++ ) {
        v = bytes[j] & 0xFF;
        hexChars[c] = hexArray[v/16];
        c++;
        hexChars[c] = hexArray[v%16];
        c++;
    }
    return new String(hexChars, 0, c); }

Answer 21

这就是我用于 MD5 哈希值的内容：

public static String getMD5(String filename)
        throws NoSuchAlgorithmException, IOException {
    MessageDigest messageDigest = 
        java.security.MessageDigest.getInstance("MD5");

    InputStream in = new FileInputStream(filename);

    byte [] buffer = new byte[8192];
    int len = in.read(buffer, 0, buffer.length);

    while (len > 0) {
        messageDigest.update(buffer, 0, len);
        len = in.read(buffer, 0, buffer.length);
    }
    in.close();

    return new BigInteger(1, messageDigest.digest()).toString(16);
}

编辑：我已经测试过了，我注意到这样尾随的零也被删除了。 但这只能发生在开始时，因此您可以与预期长度进行比较并相应地进行填充。

Answer 22

您可以在没有外部库的情况下减少编写量：

String hex = (new HexBinaryAdapter()).marshal(md5.digest(YOUR_STRING.getBytes()))

Answer 23

该解决方案不需要位移或掩码、查找表或外部库，并且大约是我能得到的最短的：

byte[] digest = new byte[16];       

Formatter fmt = new Formatter();    
for (byte b : digest) { 
  fmt.format("%02X", b);    
}

fmt.toString()

Answer 24

byte messageDigest[] = algorithm.digest();
StringBuffer hexString = new StringBuffer();
for (int i = 0; i < messageDigest.length; i++) {
    String hexByte = Integer.toHexString(0xFF & messageDigest[i]);
    int numDigits = 2 - hexByte.length();
    while (numDigits-- > 0) {
        hexString.append('0');
    }
    hexString.append(hexByte);
}

Answer 25

恕我直言，上面所有提供删除前导零的片段的解决方案都是错误的。

byte messageDigest[] = algorithm.digest();
for (int i = 0; i < messageDigest.length; i++) {
    hexString.append(Integer.toHexString(0xFF & messageDigest[i]));
}    

根据此代码片段，从字节数组中取出 8 位
迭代，转换为整数（因为 Integer.toHexString 函数采用
int 作为参数），然后将该整数转换为相应的哈希值
价值。 因此，例如，如果您有二进制 00000001 00000001，根据
在代码中，hexString 变量的十六进制值将为 0x11，而
正确的值应该是0x0101。 因此，在计算 MD5 时我们可能会得到哈希值
长度<32字节（因为缺少零）可能不满足
MD5 哈希所具有的加密独特属性。

问题的解决方案是将上面的代码片段替换为
以下片段：

byte messageDigest[] = algorithm.digest();
for (int i = 0; i < messageDigest.length; i++) {
    int temp=0xFF & messageDigest[i];
    String s=Integer.toHexString(temp);
    if(temp<=0x0F){
        s="0"+s;
    }
    hexString.append(s);
}

Answer 26

这将为一个字节提供两个字符长的字符串。

public String toString(byte b){
    final char[] Hex = new String("0123456789ABCDEF").toCharArray();
    return  "0x"+ Hex[(b & 0xF0) >> 4]+ Hex[(b & 0x0F)];
}

Answer 27

如何再次从 ascii 转换回字节数组？

我按照 Jemenake 给出的以下代码转换为 ascii。

public static String toHexString(byte[] bytes) {
    char[] hexArray = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    char[] hexChars = new char[bytes.length * 2];
    int v;
    for ( int j = 0; j < bytes.length; j++ ) {
        v = bytes[j] & 0xFF;
        hexChars[j*2] = hexArray[v/16];
        hexChars[j*2 + 1] = hexArray[v%16];
    }
    return new String(hexChars);
}

Answer 28

我的变体

    StringBuilder builder = new StringBuilder();
    for (byte b : bytes)
    {
        builder.append(Character.forDigit(b/16, 16));
        builder.append(Character.forDigit(b % 16, 16));
    }
    System.out.println(builder.toString());

它对我有用。