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C# XML到图像中的Blob字节数组_C#_Image_Bytearray_Blob - Fatal编程技术网

C# XML到图像中的Blob字节数组

C# XML到图像中的Blob字节数组,c#,image,bytearray,blob,C#,Image,Bytearray,Blob,我得到一个XML文件来生成预览,格式如下: string clean = valueC.ElementAt(0).Value.Replace("\t", string.Empty).Replace("\n", string.Empty); 预览16x16 {03F5C6D3-ABCD-4889-B3AA-C3524C62FA1C} -1 /9j/4AAQSKZJRGABAQAAAQABAD/2WBDAAUDBAQEAWUEBAQFBUGBWWIBWCHBW8LCWKMEQ8S EHEPRET

我得到一个XML文件来生成预览,格式如下:

string clean = valueC.ElementAt(0).Value.Replace("\t", string.Empty).Replace("\n", string.Empty);

预览16x16
{03F5C6D3-ABCD-4889-B3AA-C3524C62FA1C}
-1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参见E9DT4Z8G/FjxHrmla3Y3bRlRtl1iW9ztGTnlXLS47YG0nrXQa78Bda07
U7JXNM1NS9YGVHJVK8WKEKCHWAYJCUDZXGGGUB8K6V49I8X28WG+IodMkaUQFI4sQSEt1
EM5B42QOPLVCVCV5M8NKDSMW/cV3vrfXyb/AB/I+4+sVUnPCUVeyV3Z3010923krdPU/9k=

我需要将
部分转换为图像。我访问元素值的方式如下:

string clean = valueC.ElementAt(0).Value.Replace("\t", string.Empty).Replace("\n", string.Empty);
我尝试将其读入MemoryStream并转换为图像:

MemoryStream ms = new MemoryStream(blob, 0, blob.Length);
ms.Write(blob, 0, blob.Length);
Image i = Image.FromStream(ms);
通过这种方式,我在获取图像时得到“参数无效异常”。 我还尝试将其直接保存到文件中:

using (FileStream fs = new FileStream(label + ".jpg", FileMode.Create))
{
    fs.Write(blob, 0, blob.Length);
}
但当我试图打开生成的文件时,会显示一条关于文件损坏的消息

我知道编码很重要,我已经尝试过ASCII、UTF-8、UTF-7以及以下内容:

BinaryFormatter bf = new BinaryFormatter();
MemoryStream ms = new MemoryStream();
bf.Serialize(ms, clean);
ms.Seek(0, 0);
byte[] blob = ms.ToArray();
我不知道还能做什么。如果有人能帮助我,我将不胜感激


谢谢

看来您的数据是Base64格式的,所以您首先需要对其进行解码,然后将其保存为图像