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java - 十六进制字符串到字节数组的转换?

转载 作者:行者123 更新时间:2023-11-30 03:55:23 26 4
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您好,我正在尝试将十六进制字符串转换为字节数组,但我在线程“main”java.lang.NumberFormatException中收到异常作为异常:对于输入字符串:“”。

十六进制字符串:

  String sb = "0d 02 01 02 4e 52 30 39 47 30 35 36 31 38 00 03 00 fa 01 95 \n"
+ "e4 53 c0 a8 51 b1 53 a3 03 40 89 54 66 9c ee 10 01 de 9a 5c \n"
+ "22 e6 a0 2c 5d b2 69 8c 9d 79 e6 a5 41 50 c6 78 ab 31 8c 8a \n"
+ "00 72 83 53 2a e0 67 14 d0 2a 54 1b 85 00 00 76 a6 b4 64 1e \n"
+ "95 32 a1 c8 a5 90 f3 81 da 98 15 88 c7 51 54 6e 63 cb 1e 3a \n"
+ "d6 99 19 19 ef 55 6e 13 2b 91 49 89 98 2f a7 49 b8 95 20 8a \n"
+ "88 d9 cc b9 f9 0d 6d 85 e6 9e 23 e3 a5 06 7c a6 02 c6 e8 c3 \n"
+ "2a 72 0d 74 90 1d d1 29 f5 14 cf 29 5b 82 01 fa 8a 9d 14 00 \n"
+ "00 e0 50 38 ad 46 4c 38 15 89 7a b8 9c fa 11 5b f2 2e 56 a9 \n"
+ "5c e9 ad 39 0e 18 03 8e f4 8a 68 c4 14 d2 39 ad 07 d2 ae 17 \n"
+ "a0 0d f4 35 5d ed 27 4f bd 1b 0f c2 99 9d 88 06 41 ae 9f 4e \n"
+ "19 b2 88 fb 57 36 54 83 c8 23 eb c5 74 da 70 c5 94 7f 4a 07 \n"
+ "1d cb 2c 3e 53 58 77 1c ca 45 6e 37 dd 35 89 38 fd e9 a1 97 \n"
+ "2d 88 76 d1 8a 76 29 c8 bb 98 52 32 23 da 4f 6a 96 38 01 19 \n"
+ "6e 95 21 c0 e0 0a 98 0d d1 0c 75 14 0a c4 b6 d1 a0 8f 81 8c \n"
+ "d3 9d 76 1a 92 24 21 00 a8 e7 38 fc 29 9b c7 62 85 c9 dc f8 \n"
+ "f4 aa f8 a7 bb 65 89 a6 1e b4 8c a5 ab 0a 43 41 a6 f7 a0 43 \n"
+ "5e 95 54 e0 50 06 48 06 a4 c7 14 0e 28 85 86 2a 32 78 a9 5e \n"
+ "a0 90 e0 52 35 7b 12 e2 81 4b 8e 69 71 4c 60 bd 6a 74 a8 50 \n"
+ "73 53 31 c0 da 3a d0 00 cd b8 e0 74 14 a0 8c e3 bd 20 18 1c \n"
+ "75 a6 a9 cb 8d c3 9c d3 02 c4 7c 1a b8 83 8a aa 8a 59 b8 ab \n"
+ "89 c0 02 80 1e 38 a5 52 41 e2 90 0a 51 40 0f f3 1b 34 ed e1 \n"
+ "ba f5 a8 a9 71 40 12 16 e3 02 a2 75 ca 9a 78 a5 c1 34 01 45 \n"
+ "46 3a d4 8b 43 ae d9 0f bd 0b 40 89 00 14 f0 05 31 4d 48 a2 \n"
+ "81 8e db 9a 70 5a 45 19 a9 00 a6 16 1a 12 83 1e 6a 40 29 c1 \n"
+ "68 15 91 58 db 46 df 79 14 fe 15 24 71 88 d7 6a 8c 01 da a6 \n"
+ "c5 21 5a 01 22 0d 0a 2a 4b 57 00 0f 00 01 82 03 00 00 0d 0a ";
String raw_DeviceId = sb.substring(37, 1040);

byte[] b = HexStringToByteArray(raw_DeviceId);

将 HexString 转换为ByteArray

public static byte[] HexStringToByteArray(String hexStr) {
byte bArray[] = new byte[hexStr.length()/2];
for(int i=0; i<(hexStr.length()/2); i++){
byte firstNibble = Byte.parseByte(hexStr.substring(2*i,2*i+1),16); // [x,y)
byte secondNibble = Byte.parseByte(hexStr.substring(2*i+1,2*i+2),16);
int finalByte = (secondNibble) | (firstNibble << 4 ); // bit-operations only with numbers, not bytes.
bArray[i] = (byte) finalByte;
}
return bArray;
}

更新的代码:

 byte bArray[] = new byte[hexStr.length() / 2];
for (int i = 0; i < (hexStr.length() / 2); i++) {
if (hexStr.charAt(i) == ' ') {
continue;
} else {
byte firstNibble = Byte.parseByte(hexStr.substring(2 * i, 2 * i + 1), 16); // [x,y)
byte secondNibble = Byte.parseByte(hexStr.substring(2 * i + 1, 2 * i + 2), 16);
int finalByte = (secondNibble) | (firstNibble << 4); // bit-operations only with numbers, not bytes.
bArray[i] = (byte) finalByte;
}
}
return bArray;

更新:

 for (int i = 0; i < (hexStr.length() / 2); i++) {
if (hexStr.charAt(i) == ' ') {
continue;
}
byte firstNibble = Byte.parseByte(hexStr.substring(2 * i, 2 * i + 1), 16); // [x,y)
byte secondNibble = Byte.parseByte(hexStr.substring(2 * i + 1, 2 * i + 2), 16);
int finalByte = (secondNibble) | (firstNibble << 4); // bit-operations only with numbers, not bytes.
bArray[i] = (byte) finalByte;
}

更新:

1 38 00 03 00 fa 01 95 
e4 53 c0 a8 51 b1 53 a3 03 40 89 54 66 9c ee 10 01 de 9a 5c
22 e6 a0 2c 5d b2 69 8c 9d 79 e6 a5 41 50 c6 78 ab 31 8c 8a
00 72 83 53 2a e0 67 14 d0 2a 54 1b 85 00 00 76 a6 b4 64 1e
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 2
95 32 a1 c8 a5 90 f3 81 da 98 15 88 c7 51 54 6e 63 cb 1e 3a
d6 99 19 19 ef 55 6e 13 2b 91 49 89 98 2f a7 49 b8 95 20 8a
88 d9 cc b9 f9 0d 6d 85 e6 9e 23 e3 a5 06 7 at java.lang.String.substring(String.java:1907)
at java.lang.String.substring(String.java:1907)
c a6 02 c6 e8 c3
at Image.HexStringToByteArray(Image.java:28)
at Image.main(Image.java:73)
2a 72 0d 74 90 1d d1 29 f5 14 cf 29 5b 82 01 fa 8a 9d 14 00
00 e0 50 38 ad 46 4c 38 15 89 7a b8 9c fa 11 5b f2 2e 56 a9
5c e9 ad 39 0e 18 03 8e f4 8a 68 c4 14 d2 39 ad 07 d2 ae 17
a0 0d f4 35 5d ed 27 4f bd 1b 0f c2 99 9d 88 06 41 ae 9f 4e
19 b2 88 fb 57 36 54 83 c8 23 eb c5 74 da 70 c5 94 7f 4a 07
1d cb 2c 3e 53 58 77 1c ca 45 6e 37 dd 35 89 38 fd e9 a1 97
2d 88 76 d1 8a 76 29 c8 bb 98 52 32 23 da 4f 6a 96 38 01 19
6e 95 21 c0 e0 0a 98 0d d1 0c 75 14 0a c4 b6 d1 a0 8f 81 8c
d3 9d 76 1a 92 24 21 00 a8 e7 38 fc 29 9b c7 62 85 c9 dc f8
f4 aa f8 a7 bb 65 89 a6 1e b4 8c a5 ab 0a 43 41 a6 f7 a0 43
5e

更新:

public static byte[] HexStringToByteArray(String hexStr) {
System.out.println(hexStr); // log our input make sure it is what we think it is.
byte bArray[] = new byte[hexStr.length() / 3];
int index = 0;
for (String cell : hexStr.split(" ")) {
byte firstNibble = Byte.parseByte(cell.substring(0, 1), 16); // [x,y)
byte secondNibble = Byte.parseByte(cell.substring(1, 2), 16);
byte finalByte = (byte) ((secondNibble) | (firstNibble << 4)); // bit-operations only with numbers, not bytes.
bArray[index++] = finalByte;
}
return bArray;
}

最佳答案

您的字符串hexStr在连续字节之间包含空格,您需要跳过它们。一种方法是放置

if(hexStr.charAt(i) == ' ') continue;

在循环中。

当您放置继续时,您不需要使用else:

Byte bArray[] = new byte[hexStr.length()/2];  
for(int i=0; i<(hexStr.length()/2); i++){
if(hexStr.charAt(i) == ' ') continue;
byte firstNibble = Byte.parseByte(hexStr.substring(2*i,2*i+1),16); // [x,y)
byte secondNibble = Byte.parseByte(hexStr.substring(2*i+1,2*i+2),16);
int finalByte = (secondNibble) | (firstNibble << 4 ); // bit-operations only with numbers, not bytes.
bArray[i] = (byte) finalByte;
}

关于java - 十六进制字符串到字节数组的转换?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23354999/

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