16 진수 문자열을 이미지로 변환

Question

iPad 용 FormsConnect 응용 프로그램을 사용하여 서명을 XML로 내 보냅니다. 이 코드를 사용해서 이미지에 다시이 진수 문자열로 변환하기 위해 노력하고있어 이제

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

: 다음 앱에서 출력 될 때

Bitmap newImage = new Bitmap(new MemoryStream(StringToByteArray(hex))); 

public static byte[] StringToByteArray(string hex) 
    { 
     hex = hex.Substring(2, hex.Length - 2); 
     if (hex.Length % 2 != 0) 
     { 
      throw new ArgumentException(String.Format(CultureInfo.InvariantCulture, "The binary key cannot have an odd number of digits: {0}", hex)); 
     } 

     byte[] HexAsBytes = new byte[hex.Length/2]; 

     for (int index = 0; index < HexAsBytes.Length; index++) 
     { 
      string byteValue = hex.Substring(index * 2, 2); 
      HexAsBytes[index] = byte.Parse(byteValue, NumberStyles.HexNumber, CultureInfo.InvariantCulture); 
     } 

     return HexAsBytes; 
    } 

을하지만 매개 변수를 점점 유효하지 않습니다 유지 스트림을 비트 맵에로드하려고 할 때. 나는 앱의 개발자에게 문제를 제기하고, 여기에 자신의 반응이다 :

친애하는 알렉스

데이터는 바이너리, 진수 형식으로 저장됩니다. 다음은 이진 데이터를 읽고이를 jpg로 다시 쓰는 방법에 대한 지침입니다.

는 byteData [I]는 = (자리 SHL 4 나 2 *]) +

위의 숫자는 이미 0 내지 15의 숫자로 변환된다고 가정 (숫자 [I 2 + 1 *]). 더 큰 덩어리 (정수 등)로 메모리에 작성하여 스크립트를 최적화하려는 경우 엔디안을 고려해야합니다. 모든 주요 CPU 아키텍처는 리틀 엔디안 (little endian)입니다. 즉 숫자가 실제 값과 반대 순서로 배열 된 바이트로 저장됩니다 (예 : 0x1234가 메모리에 0x3412로 저장 됨). 사람이 나를 도울 수 있다면

1. Python Script: 
iterate over each photo attached to image field in FormConnect xml file 
data = R[i] 

to strip '0x' off front of each blob 
a = data[2:] 
iterate over each byte, write as binary int to array 
for i in range(len(a)/2): 
       b = (int(bin(int(a[i*2],16)),2)<<4) + int(bin(int(a[i*2+1],16)),2) 
       data.append(b) 
write array to binary file 

F = file(outFile,'wb') 
data.tofile(F) 
F.close() 


PHP Script: 
$pic = ""; 
$field = substr($field,2); 
header("Content-Type: image/jpeg"); 
$pic = pack("H*", $field); 
echo $pic; 

, 나는 매우 감사하게 될 것입니다 :

아래 양식에서 BLOB 데이터를 추출하기위한 두 개의 서로 다른 스크립트가 있습니다. 친절한 답변, 알렉스

Answer 1

문제는 16 진수 변환 루틴이 아닙니다. 원래 데이터가 올바르게 인코딩되지 않았기 때문입니다. 그 안에있는 모든 0을 기반으로, 난 그냥 원시 비트 맵이라고 말하고 싶지만! 내 생각 엔 그 데이터를 생성 한 프로그램의 제작자가 비트 맵을 출력하기 전에 특정 파일 형식으로 인코딩하는 것을 잊어 버렸다는 것입니다.