java 以字符串获取的数组，怎么转成16位字符串？

String msg="[104,-62,57,-106,119,81,32,71,102,-48,-71,38,0,8,8,0,0,-18,-18,-18,-18,-18,-18,-18,-18,-18,-18,19,23,22,18,34,36,32,70,52,69,48,48,8,8,-12,-4,40,0,0,0,1,-18,-18,-18,-94,-64,22]";
这个是获取的字符串,该怎么转成16进制字符串呢？
或者转化成

byte msgs[]=
{104,-62,57,-106,119,81,32,71,102,-48,-71,38,0,8,8,0,0,-18,-18,-18,-18,-18,-18,-18,-18,-18,-18,19,23,22,18,34,36,32,70,52,69,48,48,8,8,-12,-4,40,0,0,0,1,-18,-18,-18,-94,-64,22}
也可以
最新回答
︷决戰紫禁之巓︷

2024-05-02 04:40:48

你可以使用以下步骤将字符串 msg 转换为16进制字符串：

  1. 将字符串 msg 转换为字节数组，可以使用 Arrays.toString(msg.getBytes())

  2. 将字节数组中每个字节转换为 16 进制字符串，可以使用 Integer.toHexString(byteValue)

  3. 将转换后的字符串拼接起来得到最终的16进制字符串

    例如：

    byte[] bytes = msg.getBytes();

    StringBuilder hexString = new StringBuilder();

    for (byte b : bytes) {

    hexString.append(Integer.toHexString(b & 0xff));

    }

    String result = hexString.toString();

  4. 注意：转换后的字符串可能会有一些前导0,如果需要去掉可以使用 string.replaceFirst("^0+(?!$)", "")

假丶惆怅

2024-05-02 20:48:14

可以使用Java的Integer.toHexString(int)方法将int类型转换成16进制字符串。
String hexString = "";
for (int i = 0; i < msg.length; i++) {
int num = Integer.parseInt(msg[i]);
String hex = Integer.toHexString(num);
hexString += hex;
}
System.out.println(hexString);
陌然淺笑

2024-05-02 00:28:59

方法一：

String[] arr = msg.split(",");
StringBuilder sb = new StringBuilder();
for (String str : arr) {
int value = Integer.parseInt(str.trim());
sb.append(Integer.toHexString(value)).append(" ");
}
String hexStr = sb.toString();
System.out.println(hexStr);
方法二：

String[] arr = msg.split(",");
StringBuffer sb = new StringBuffer();
for (int i = 0; i < arr.length; i++) {
String hexStr = Integer.toHexString(Integer.valueOf(arr[i]));
sb.append(hexStr).append(" ");
}
System.out.println(sb.toString());
空城仅有旧梦在

2024-05-02 16:10:46

String[] strArray = new String[]{"a","b","c"};StringBuilder sb = new StringBuilder();for (String str : strArray) { sb.append(String.format("%016x", new BigInteger(1, str.getBytes())));}String hexStr = sb.toString();System.out.println(hexStr);

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