Camera Remote API

Question

I am trying to use liveview through the Camera Remote SDK in order to connect a SONY QX1 lensless camera to a PC over the wireless link. All of the APIs I have tested so far work great except the liveview. I have not been able to find a viewer app that allows decoding of the jpg stream. Does one exist? I would prefer not to write this myself. I am currently using VB in Visual.net to communicate with the camera and can transfer files and control the camera just fine. I could just use a bit of help with the liveview process.

I can capture the stream in Chrome but it will not display the file it just wants to download the stream... So far no other viewer seems to be able to parse the stream.

Any help would be appreciated. I hope I have been clear in describing the problem... I like the camera.

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As per the comment, here is some code that works with the camera. It is written in VB6 but presumably will work with Visual.Net:

Hi,

Below is an example of code I have used for communicating with the camera using VB6 and MSXML2.xmlhttp. It works fine for all commands: I can take pictures, start and stop movies, etc. using the API commands in the Sony SDK. The code below takes a still picture with the QX1 and stores it on the internal memory card.

For liveview I enter the URL for the liveview stream (http://192.168.122.1:8080/liveview/liveviewstream) into Chrome and the browser starts to download the file but does not display the stream, just stores it in the download file location.

I am looking for a procedure for parsing and displaying the stream in a web browser or picture box.

Tom

Dim sURL As String
sURL = "http://192.168.122.1:8080/sony/camera"

Dim xmlhttp As New MSXML2.xmlhttp



Dim strXML() As Byte
strXML = "{""method"": ""actTakePicture"", ""params"":[ ], ""id"":  1, ""version"": ""1.0""}" & vbCrLf
strXML = StrConv(strXML, vbFromUnicode)

'Start of code to use HTTP POST
xmlhttp.open "POST", sURL, False
xmlhttp.setRequestHeader "Content-type:json;charst=utf-8" & vbCrLf, ""

xmlhttp.send (strXML)  '
Text1.Text = xmlhttp.responsetext