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Home/ Questions/Q 798941
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T23:02:55+00:00 2026-05-14T23:02:55+00:00

I am getting a XML File to generate a preview, in a format like

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I am getting a XML File to generate a preview, in a format like this:

<PARAM>
<LABEL>Preview 16x16</LABEL>
<ID>{03F5C6D3-ABCD-4889-B3AA-C3524C62FA1C}</ID>
<LAYER>-1</LAYER>
<VALUE>
<BLOB>
/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAUDBAQEAwUEBAQFBQUGBwwIBwcHBw8LCwkMEQ8S
EhEPERETFhwXExQaFRERGCEYGh0dHx8fExciJCIeJBweHx7/2wBDAQUFBQcGBw4ICA4eFBEU
Hh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh4eHh7/wAAR
CAAOABADASIAAhEBAxEB/8QAHwAAAQUBAQEBAQEAAAAAAAAAAAECAwQFBgcICQoL/8QAtRAA
AgEDAwIEAwUFBAQAAAF9AQIDAAQRBRIhMUEGE1FhByJxFDKBkaEII0KxwRVS0fAkM2JyggkK
FhcYGRolJicoKSo0NTY3ODk6Q0RFRkdISUpTVFVWV1hZWmNkZWZnaGlqc3R1dnd4eXqDhIWG
h4iJipKTlJWWl5iZmqKjpKWmp6ipqrKztLW2t7i5usLDxMXGx8jJytLT1NXW19jZ2uHi4+Tl
5ufo6erx8vP09fb3+Pn6/8QAHwEAAwEBAQEBAQEBAQAAAAAAAAECAwQFBgcICQoL/8QAtREA
AgECBAQDBAcFBAQAAQJ3AAECAxEEBSExBhJBUQdhcRMiMoEIFEKRobHBCSMzUvAVYnLRChYk
NOEl8RcYGRomJygpKjU2Nzg5OkNERUZHSElKU1RVVldYWVpjZGVmZ2hpanN0dXZ3eHl6goOE
hYaHiImKkpOUlZaXmJmaoqOkpaanqKmqsrO0tba3uLm6wsPExcbHyMnK0tPU1dbX2Nna4uPk
5ebn6Onq8vP09fb3+Pn6/9oADAMBAAIRAxEAPwDP+IXxH+Ilr4/1DRrLxZqUrjUTbMLKdtsz
eZgLCgHyDHy4GSeee9dT4Z8G/FjxHrmla3Y3bRlRtl1iW9ztGTnlXLS47YG0nrXQa78Bda07
U7jXNM1nS9yGVhJLHJvk8wkEkchWAYjcuDzxggGuB8K6v49i8X28Wg+IodMkaUQFI4sQSEt1
eM5B42qOPlVVVcAV5M8NKdSMW/cV3vrfXyb/AB/I+4+sVUnPCUVeyV3Z3010923krdPU/9k=
</BLOB>
</VALUE>
</PARAM>

I need to convert the <BLOB> section into an Image. I access the Element Value like this:

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

I’ve tried to read it into a MemoryStream and convert to Image:

MemoryStream ms = new MemoryStream(blob, 0, blob.Length);
ms.Write(blob, 0, blob.Length);
Image i = Image.FromStream(ms);

In this way I get “Parameter not valid exception” at getting the Image.
I’ve also tried to save it directly into a File:

using (FileStream fs = new FileStream(label + ".jpg", FileMode.Create))
{
    fs.Write(blob, 0, blob.Length);
}

But when I try to open the generated file, displays a message about damage in it.

I know that encoding is important, I’ve already tried ASCII, UTF-8, UTF-7 and this:

BinaryFormatter bf = new BinaryFormatter();
MemoryStream ms = new MemoryStream();
bf.Serialize(ms, clean);
ms.Seek(0, 0);
byte[] blob = ms.ToArray();

I dont know what else to do. I’ll appreciate if somebody can help me.

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T23:02:56+00:00Added an answer on May 14, 2026 at 11:02 pm

    It seems your data is in Base64 format so you first need to decode it and then save it as an image.

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