The numbers that you encode into hexadecimal must represent some encoding of the characters, such as UTF-8. So first convert the String to a byte[] representing the string in that encoding, then convert each byte to hexadecimal.
public static String hexadecimal(String input, String charsetName) throws UnsupportedEncodingException {
if (input == null) throw new NullPointerException();
return asHex(input.getBytes(charsetName));
}
private static final char[] HEX_CHARS = "0123456789abcdef".toCharArray();
public static String asHex(byte[] buf)
{
char[] chars = new char[2 * buf.length];
for (int i = 0; i < buf.length; ++i)
{
chars[2 * i] = HEX_CHARS[(buf[i] & 0xF0) >>> 4];
chars[2 * i + 1] = HEX_CHARS[buf[i] & 0x0F];
}
return new String(chars);
}
byte[] bytes = string.getBytes(CHARSET); // you didn't say what charset you wanted
BigInteger bigInt = new BigInteger(bytes);
String hexString = bigInt.toString(16); // 16 is the radix
All answers based on String.getBytes() involve 编码 your string according to a Charset. You don't necessarily get the hex value of the 2-byte 角色 that make up your string. If what you actually want is the equivalent of a hex viewer, then you need to access the chars directly. Here's the function that I use in my code for debugging Unicode issues:
static String stringToHex(String string) {
StringBuilder buf = new StringBuilder(200);
for (char ch: string.toCharArray()) {
if (buf.length() > 0)
buf.append(' ');
buf.append(String.format("%04x", (int) ch));
}
return buf.toString();
}
public static void main(String... args){
String str = "Hello! This is test string.";
char ch[] = str.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < ch.length; i++) {
sb.append(Integer.toHexString((int) ch[i]));
}
System.out.println(sb.toString());
}