Guest
14 years ago
Hi, I am trying to send a JPEG (only 7k) via MMSSend, when i receive the message on my mobile the JPEG is corrupt only 1k?

Code below:

String _fileName = "pic.jpg";
String _file = "c:\\Pictures\\" + _fileName;

// -------------------------------------------------------------------------------
// SEND MESSAGE
IntelliSoftware.MMSMessage mmsMsg = new IntelliSoftware.MMSMessage();
mmsMsg.Subject = "Test";

IntelliSoftware.MessagePart msgPart1 = new IntelliSoftware.MessagePart();
msgPart1.Filename = _file;
msgPart1.SourceFilename = _fileName;
msgPart1.Text = "Picture";
mmsMsg.MessageParts.Add(msgPart1);

IntelliSoftware.MessagePart msgPart2 = new IntelliSoftware.MessagePart();
msgPart2.Filename = "text.txt";
msgPart2.Text = "Test Text";
mmsMsg.MessageParts.Add(msgPart2);

IntelliSoftware.IntelliSMS objIntelliSMS = new IntelliSoftware.IntelliSMS();
objIntelliSMS.Username = "abc";
objIntelliSMS.Password = "xyz";

StringCollection toList = new StringCollection();
toList.Add(12345678);

IntelliSoftware.SendStatusCollection statusCollection;
statusCollection = objIntelliSMS.SendMMSMessage(toList, mmsMsg);
// -------------------------------------------------------------------------------

Any ideas?

Thanks Andy
Guest
14 years ago
Hi, Just to add an update to my problem.

I moved away from specifying a SourceFilename, and instead used the SourceStream property as below:

FileStream fs = new FileStream(_fileFullName, FileMode.Open);
msgPart1.SourceStream = fs;

This now uploads perfectly to your server, I can see the message in my "Sent" items mailbox and the image is correct.

Does this mean there is a bug in the SourceFilename property?

Furthermore, I am still not receiving the Image correctly onto my phone (HTC HD2) as I am getting the message
"Unable to download message from xxxxxxxxxx"

So, to conclude, I have successfully uploaded the graphic to your server correctly. I can now verify this in my "Sent" items mailbox.
I still cannot download the graphic from your server onto my phone.

Regards

Andy
Support
14 years ago
Regarding your first issue, it looks like you have switched the parameters, please try:

msgPart1.Filename = _fileName;
msgPart1.SourceFilename = _file;

'SourceFilename' needs to contain the full path and filename. 'Filename' just the filename.