将十六进制字符串转换为图像文件[重复]
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【中文标题】将十六进制字符串转换为图像文件[重复]【英文标题】:Converting a Hex String to an image file [duplicate] 【发布时间】:2012-01-27 07:04:03 【问题描述】:我需要从接收到的 xml 文件中读取带有前导零的十六进制字符串,该字符串表示 JPEG 文件,并将其保存为图像文件。图像数据看起来像
0000005000000050FF191818FF151715FF111413FF0E1...........................FF2A2322FF292221
在xml文件中标签之间的字符串长度为51216我将Photo标签之间的十六进制数据作为字符串读取并将其转换为字节[],并使用FileOutputStream
我正在写入文件.但是当我尝试打开图像文件时,它告诉“文件似乎已损坏、损坏或文件太大”我尝试了很多方法来保存图像,但都没有成功。我列出了下面使用的方法。请帮我解决这个问题。
String photo="0000005000000050FF191818FF15"; //this is just a sample.The photo String actually contains the full Hex String which is 51216 long
//METHOD 1
String[] v = photo.split(" ");
byte[] arr = new byte[v.length];
int x = 0;
for(String val: v)
arr[x++] = Integer.decode("0x" + val).byteValue();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(arr);
fos.flush();
fos.close();
//METHOD 2
byte[] arr = new byte[photo.length()/2];
for ( int start = 0; start < photo.length(); start += 2 )
String thisByte = photo.substring(start, start+2);
arr[start/2] = Byte.parseByte(thisByte, 16);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(arr);
fos.flush();
fos.close();
//METHOD 3
if ((photo.length() % 2) != 0)
throw new IllegalArgumentException("Input string must contain an even number of characters");
final byte result[] = new byte[photo.length()/2];
final char enc[] = photo.toCharArray();
for (int x = 0; x < enc.length; x += 2)
StringBuilder curr = new StringBuilder(2);
curr.append(enc[x]).append(enc[x + 1]);
result[x/2] = (byte) Integer.parseInt(curr.toString(), 16);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(result);
fos.flush();
fos.close();
//METHOD 4
byte result[] = new byte[photo.length()/2];
char enc[] = photo.toUpperCase().toCharArray();
StringBuffer curr;
for (int x = 0; x < enc.length; x += 2)
curr = new StringBuffer("");
curr.append(String.valueOf(enc[x]));
curr.append(String.valueOf(enc[x + 1]));
result[x] = (byte) Integer.parseInt(curr.toString(), 16);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(result);
fos.flush();
fos.close();
//METHOD 5
int len = photo.length();
byte[] data = new byte[len / 2];
for (int x = 0; x < len; x += 2)
data[x / 2] = (byte) ((Character.digit(photo.charAt(x), 16) << 4)
+ Character.digit(photo.charAt(x+1), 16));
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(data);
fos.flush();
fos.close();
//METHOD 6
byte[] bytes=new BigInteger(photo, 16).toByteArray();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(bytes);
fos.flush();
fos.close();
//METHOD 7
byte[] bytes =DatatypeConverter.parseHexBinary(photo);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(bytes);
fos.flush();
fos.close();
//METHOD 8
HexBinaryAdapter adapter = new HexBinaryAdapter();
byte[] bytes = adapter.unmarshal(photo);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.png");
fos.write(bytes);
fos.flush();
fos.close();
//METHOD 9
byte data[] = new byte[photo.length()/2];
for(int x=0;i < photo.length();x+=2)
data[x/2] = (Integer.decode("0x"+photo.charAt(x)+photo.charAt(x+1))).byteValue();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(data);
fos.flush();
fos.close();
//METHOD 10
byte[] data = new byte[photo.length()/2];
for (int x=0;i<photo.length()/2;x++)
data[x] = (Integer.decode(
"0x"+photo.substring(x*2, (x+1)*2))).byteValue();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(data);
fos.flush();
fos.close();
//METHOD 11
String hexVal ="0000005000000050FF";
//String hexVal = "0123456789ABCDEF";
byte[] out = new byte[photo.length() / 2];
int n = photo.length();
for( int x = 0; x < n; x += 2 )
int hn = hexVal.indexOf( photo.charAt( x ) );
int ln = hexVal.indexOf( photo.charAt( x + 1 ) );
out[x/2] = (byte)( ( hn << 4 ) | ln );
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(out);
fos.flush();
fos.close();
//METHOD 12
byte[] array=photo.getBytes();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(array);
fos.flush();
fos.close();
//METHOD 13
byte[] array=photo.getBytes();
byte[] bytes = Base64.decode(array);
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(bytes);
fos.flush();
fos.close();
//METHOD 14
byte[] array=photo.getBytes();
Charset csets = Charset.forName("UTF-8");
ByteBuffer bb=ByteBuffer.wrap(array);
csets.decode(bb);
bb.rewind();
byte[] array1=bb.array();
FileOutputStream fos=new FileOutputStream("D:/Images/image6.jpg");
fos.write(array1);
fos.flush();
fos.close();
【问题讨论】:
如何“代表一个 JPG 文件”?它不是原始 JPG,也不是 JFIF 或 BASE64 或任何东西。 看起来它是十六进制编码的,但据我了解,JPG 不应该以 0x00、0x00 开头,而是以 0xff、0xd8 开头。见en.wikipedia.org/wiki/JPEG#Syntax_and_structure 嗨@Nick。你成功了吗?? 【参考方案1】:最短的方法可能是这样。
String photo = "0000005000000050FF191818FF151715FF111413FF0E100FF2A2322FF292221";
// adds a dummy byte at the start to avoid truncation of leading zeros.
byte[] bytes = new BigInteger("10" + photo, 16).toByteArray();
System.out.println(Arrays.toString(bytes));
打印
[1, 0, 0, 0, 5, 0, 0, 0, 5, 15, -15, -111, -127, -113, -15, 81, 113, 95, -15, 17, 65、63、-16、-31、0、-1、42、35、34、-1、41、34、33]
【讨论】:
嗨彼得,这也不起作用 @Nick 以什么方式?您是否尝试过将生成的字节数组写入文件进行测试? @Nick,不工作是什么意思?它为我产生了预期的结果。 是的,我使用 FileOutputSram byte[] bytes = new BigInteger("10" + photo, 16).toByteArray(); 将字节数组写入文件FileOutputStream fos=new FileOutputStream("D://Images/image.jpg"); fos.write(字节); fos.flush(); fos.close(); 您需要删除“虚拟字节”;例如fos.write(bytes, 1, bytes.length-1);
【参考方案2】:
public class test
static String HEX_STRING = "0123456789ABCDEF";
public static byte[] convertHexadecimal2Binary(byte[] hex)
int block = 0;
byte[] data = new byte[hex.length / 2];
int index = 0;
boolean next = false;
for (int i = 0; i < hex.length; i++)
block <<= 4;
int pos = HEX_STRING.indexOf(Character.toUpperCase((char) hex[i]));
if (pos > -1)
block += pos;
if (next)
data[index] = (byte) (block & 0xff);
index++;
next = false;
else
next = true;
return data;
public static void main(String args[])
String line = "";
String line_final = "";
try
String sCurrentLine;
BufferedReader br = new BufferedReader(new FileReader("D:\\test.txt"));//test.txt hex code string
DataOutputStream os = new DataOutputStream(new FileOutputStream("D:\\mohit.jpg"));
while ((sCurrentLine = br.readLine()) != null)
line = StringUtils.deleteWhitespace(sCurrentLine);
byte[] temp = convertHexadecimal2Binary(line.getBytes());
os.write(temp);
os.close();
br.close();
catch (Exception e)
e.printStackTrace();
【讨论】:
【参考方案3】:当您检索文件时,将在十六进制字符串的开头附加一个空格,因此请删除该空格并将其存储在另一个字节数组中,它可以正常工作:
这里是删除起始空间的代码
byte a3[] = new BigInteger(str, 16).toByteArray();
byte a[] = new byte[a3.length - 1];
for (int i = 1; i < a3.length; i++)
a[i - 1] = a3[i];
这里a3
包含检索到的字节数据;a1
包含实际的字节流
【讨论】:
【参考方案4】:方法 2 看起来是正确的(尚未检查所有其他方法)——您的问题可能出在其他地方。您确定从 XML 中提取的字符串是完整的吗?您使用的是哪个解析器?也许它返回多个部分的长字符串(我认为这可能是 SAX 解析器的情况),而您只提取第一部分?
以下是我将如何实现解码部分(通过substring
、BigInteger
、char[]
等避免不必要的额外分配;但为了性能,您可能希望使用 BufferedOutputStream):
String photo = "0000005000000050FF191818FF15";
FileOutputStream fos = new FileOutputStream("D:/Images/image6.jpg");
for (int i = 0; i < photo.length; i += 2)
int byte = Character.digit(photo.charAt(i), 16) * 16 +
Character.digit(photo.charAt(i + 1), 16);
fos.write(byte);
fos.close();
【讨论】:
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