URGENT, tricky question (BitBlt experts especially)

[VB Programmer]

I am using the BitBlt operation to capture various controls into jpegs.
It's almost like a screen capture, but of just the control. (This is a
VB.NET application.)

Because of BitBlt limitations I know that the application has to always be
on top.

The problem: I am running this application on a Windows Server 2003 PC. The
server is in another state (North Carolina). I am in Florida. I access the
server over Termincal Server /Remote Desktop. When I disconnect my Remote
Desktop (with the application still running) ALL THE GENERATED IMAGES ARE
BLACK! If I reconnect with Remote Desktop it works PERFECTLY. In other
words it's like using my remote display to draw the jpegs!

1. Any ideas how to overcome this tricky problem?
2. Any alternatives to BitBlt? (I have an ActiveX gauge control (like a
speedometer) and need to capture just the control to a jpeg image.)

Thanks.

[Ken Tucker [MVP]]

Hi,

Post your code

Ken
---------------------
"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:%2******** ********@TK2MSF TNGP10.phx.gbl. ..

I am using the BitBlt operation to capture various controls into jpegs.
It's almost like a screen capture, but of just the control. (This is a
VB.NET application.)

Because of BitBlt limitations I know that the application has to always be
on top.

The problem: I am running this application on a Windows Server 2003 PC.
The
server is in another state (North Carolina). I am in Florida. I access
the
server over Termincal Server /Remote Desktop. When I disconnect my Remote
Desktop (with the application still running) ALL THE GENERATED IMAGES ARE
BLACK! If I reconnect with Remote Desktop it works PERFECTLY. In other
words it's like using my remote display to draw the jpegs!

1. Any ideas how to overcome this tricky problem?
2. Any alternatives to BitBlt? (I have an ActiveX gauge control (like a
speedometer) and need to capture just the control to a jpeg image.)

Thanks.

[VB Programmer]

Here's the portion dealing with that...

Public Declare Function BitBlt Lib "gdi32" ( _
ByVal hDestDC As IntPtr, _
ByVal x As Integer, _
ByVal y As Integer, _
ByVal nWidth As Integer, _
ByVal nHeight As Integer, _
ByVal hSrcDC As IntPtr, _
ByVal xSrc As Integer, _
ByVal ySrc As Integer, _
ByVal dwRop As Integer _
) As Integer

Private Sub CaptureGauge(By Val strFileName As String, ByVal IsKnob As
Boolean)
Try
Dim MyBitmap As Bitmap
Dim MyForm As frmMain

If IsKnob Then
MyBitmap = CaptureControl( MyFrmMain.panKn ob)
Else
MyBitmap = CaptureControl( MyFrmMain.panSl ider)
End If
MyBitmap.Save(s trFileName, Imaging.ImageFo rmat.Jpeg)

Catch ex As Exception
HandleError(ex, "CaptureGau ge")
End Try
End Sub

Public Function CaptureControl( ByVal c As Control) As Bitmap
Dim bmp As Bitmap
Dim gDest, gSource As Graphics
Dim hdcSource, hdcDest As IntPtr
Const SRCCOPY As Integer = &HCC0020
bmp = New Bitmap(c.Width, c.Height)

gSource = c.CreateGraphic s
Try
gDest = Graphics.FromIm age(bmp)
Try
hdcSource = gSource.GetHdc
Try
hdcDest = gDest.GetHdc
Try
BitBlt( _
hdcDest, 0, 0, _
c.Width, c.Height, _
hdcSource, 0, 0, SRCCOPY _
)
Finally
gDest.ReleaseHd c(hdcDest)
End Try
Finally
gSource.Release Hdc(hdcSource)
End Try
Finally
gDest.Dispose()
End Try
Finally
gSource.Dispose ()
End Try
Return bmp
End Function

"Ken Tucker [MVP]" <vb***@bellsout h.net> wrote in message
news:eR******** ******@TK2MSFTN GP11.phx.gbl...

Hi,

Post your code

Ken
---------------------
"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:%2******** ********@TK2MSF TNGP10.phx.gbl. ..

I am using the BitBlt operation to capture various controls into jpegs.
It's almost like a screen capture, but of just the control. (This is a
VB.NET application.)

Because of BitBlt limitations I know that the application has to always be on top.

The problem: I am running this application on a Windows Server 2003 PC.
The
server is in another state (North Carolina). I am in Florida. I access
the
server over Termincal Server /Remote Desktop. When I disconnect my Remote Desktop (with the application still running) ALL THE GENERATED IMAGES ARE BLACK! If I reconnect with Remote Desktop it works PERFECTLY. In other
words it's like using my remote display to draw the jpegs!

1. Any ideas how to overcome this tricky problem?
2. Any alternatives to BitBlt? (I have an ActiveX gauge control (like a speedometer) and need to capture just the control to a jpeg image.)

Thanks.

[Ken Tucker [MVP]]

Hi,

Are you getting an error ?

Declare Function GetLastError Lib "kernel32" Alias "GetLastErr or" () As
Integer

Private Declare Function FormatMessage Lib "kernel32" Alias "FormatMessageA "
_

(ByVal dwFlags As Integer, ByRef lpSource As Object, ByVal dwMessageId As
Integer, _

ByVal dwLanguageId As Integer, ByVal lpBuffer As String, ByVal nSize As
Integer, _

ByRef Arguments As Integer) As Integer

Private Const FORMAT_MESSAGE_ FROM_SYSTEM As Short = &H1000S

Private Const LANG_NEUTRAL As Short = &H0S

Dim intReturn As Integer

intReturn = BitBlt( _

hdcDest, 0, 0, w, h, hdcSrc, 0, 0, SRCCOPY)

If intReturn = 0 Then

Dim Buffer As String = Space(255)

FormatMessage(F ORMAT_MESSAGE_F ROM_SYSTEM, 0, GetLastError, _

LANG_NEUTRAL, Buffer, 255, 0)

MessageBox.Show (Buffer)

End If

Ken

---------------------------

"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:OH******** ******@tk2msftn gp13.phx.gbl...

Here's the portion dealing with that...

Public Declare Function BitBlt Lib "gdi32" ( _
ByVal hDestDC As IntPtr, _
ByVal x As Integer, _
ByVal y As Integer, _
ByVal nWidth As Integer, _
ByVal nHeight As Integer, _
ByVal hSrcDC As IntPtr, _
ByVal xSrc As Integer, _
ByVal ySrc As Integer, _
ByVal dwRop As Integer _
) As Integer

Private Sub CaptureGauge(By Val strFileName As String, ByVal IsKnob As
Boolean)
Try
Dim MyBitmap As Bitmap
Dim MyForm As frmMain

If IsKnob Then
MyBitmap = CaptureControl( MyFrmMain.panKn ob)
Else
MyBitmap = CaptureControl( MyFrmMain.panSl ider)
End If
MyBitmap.Save(s trFileName, Imaging.ImageFo rmat.Jpeg)

Catch ex As Exception
HandleError(ex, "CaptureGau ge")
End Try
End Sub

Public Function CaptureControl( ByVal c As Control) As Bitmap
Dim bmp As Bitmap
Dim gDest, gSource As Graphics
Dim hdcSource, hdcDest As IntPtr
Const SRCCOPY As Integer = &HCC0020
bmp = New Bitmap(c.Width, c.Height)

gSource = c.CreateGraphic s
Try
gDest = Graphics.FromIm age(bmp)
Try
hdcSource = gSource.GetHdc
Try
hdcDest = gDest.GetHdc
Try
BitBlt( _
hdcDest, 0, 0, _
c.Width, c.Height, _
hdcSource, 0, 0, SRCCOPY _
)
Finally
gDest.ReleaseHd c(hdcDest)
End Try
Finally
gSource.Release Hdc(hdcSource)
End Try
Finally
gDest.Dispose()
End Try
Finally
gSource.Dispose ()
End Try
Return bmp
End Function

"Ken Tucker [MVP]" <vb***@bellsout h.net> wrote in message
news:eR******** ******@TK2MSFTN GP11.phx.gbl...

Hi,

Post your code

Ken
---------------------
"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:%2******** ********@TK2MSF TNGP10.phx.gbl. ..

>I am using the BitBlt operation to capture various controls into jpegs.
> It's almost like a screen capture, but of just the control. (This is a
> VB.NET application.)
>
> Because of BitBlt limitations I know that the application has to always be > on top.
>
> The problem: I am running this application on a Windows Server 2003 PC.
> The
> server is in another state (North Carolina). I am in Florida. I
> access
> the
> server over Termincal Server /Remote Desktop. When I disconnect my Remote > Desktop (with the application still running) ALL THE GENERATED IMAGES ARE > BLACK! If I reconnect with Remote Desktop it works PERFECTLY. In
> other
> words it's like using my remote display to draw the jpegs!
>
> 1. Any ideas how to overcome this tricky problem?
> 2. Any alternatives to BitBlt? (I have an ActiveX gauge control (like a > speedometer) and need to capture just the control to a jpeg image.)
>
> Thanks.
>
>

[Tom Shelton]

On 2004-02-07, Ken Tucker [MVP] <vb***@bellsout h.net> wrote:

Hi,

Are you getting an error ?

Declare Function GetLastError Lib "kernel32" Alias "GetLastErr or" () As
Integer

Private Declare Function FormatMessage Lib "kernel32" Alias "FormatMessageA "
_

(ByVal dwFlags As Integer, ByRef lpSource As Object, ByVal dwMessageId As
Integer, _

ByVal dwLanguageId As Integer, ByVal lpBuffer As String, ByVal nSize As
Integer, _

ByRef Arguments As Integer) As Integer

Private Const FORMAT_MESSAGE_ FROM_SYSTEM As Short = &H1000S

Private Const LANG_NEUTRAL As Short = &H0S

Dim intReturn As Integer

intReturn = BitBlt( _

hdcDest, 0, 0, w, h, hdcSrc, 0, 0, SRCCOPY)

If intReturn = 0 Then

Dim Buffer As String = Space(255)

FormatMessage(F ORMAT_MESSAGE_F ROM_SYSTEM, 0, GetLastError, _

LANG_NEUTRAL, Buffer, 255, 0)

MessageBox.Show (Buffer)

End If

Ken

---------------------------

"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:OH******** ******@tk2msftn gp13.phx.gbl...

Ken,

Using GetLastError in VB.NET is not a good idea - for the same reasons
as in VB.CLASSIC. The runtime makes API calls and the value you get
from GetLastError may not be correct...

To get API error returns you should use Marshal.GetLast Win32Error().
And the calls to format message are not needed either. Just create an
instance of System.Componen tModel.Win32Exc eption and then look at the
..Message property...

' BitBlt code...
If intRetun = 0 Then
Dim ex As New Win32Exception( Marshal.GetLast Win32Error())
MessageBox.Show (ex.Message)
End If

--
Tom Shelton [MVP]
Powered By Gentoo Linux 1.4
Of all men's miseries, the bitterest is this:
to know so much and have control over nothing.
-- Herodotus

[Armin Zingler]

"Ken Tucker [MVP]" <vb***@bellsout h.net> schrieb

Are you getting an error ?

Untested: I think we could reproduce the problem by pressing Win+L and
having the app make a snapshot meanwhile...
--
Armin

[Ken Tucker [MVP]]

Hi,

Much better method.

Ken
------------------
"Tom Shelton" <to*@mtogden.co m> wrote in message
news:%2******** *******@TK2MSFT NGP11.phx.gbl.. .

On 2004-02-07, Ken Tucker [MVP] <vb***@bellsout h.net> wrote:

Hi,

Are you getting an error ?

Declare Function GetLastError Lib "kernel32" Alias "GetLastErr or" () As
Integer

Private Declare Function FormatMessage Lib "kernel32" Alias
"FormatMessageA "
_

(ByVal dwFlags As Integer, ByRef lpSource As Object, ByVal dwMessageId As
Integer, _

ByVal dwLanguageId As Integer, ByVal lpBuffer As String, ByVal nSize As
Integer, _

ByRef Arguments As Integer) As Integer

Private Const FORMAT_MESSAGE_ FROM_SYSTEM As Short = &H1000S

Private Const LANG_NEUTRAL As Short = &H0S

Dim intReturn As Integer

intReturn = BitBlt( _

hdcDest, 0, 0, w, h, hdcSrc, 0, 0, SRCCOPY)

If intReturn = 0 Then

Dim Buffer As String = Space(255)

FormatMessage(F ORMAT_MESSAGE_F ROM_SYSTEM, 0, GetLastError, _

LANG_NEUTRAL, Buffer, 255, 0)

MessageBox.Show (Buffer)

End If

Ken

---------------------------

"VB Programmer" <gr*********@ go-intech.com> wrote in message
news:OH******** ******@tk2msftn gp13.phx.gbl...

Ken,

Using GetLastError in VB.NET is not a good idea - for the same reasons
as in VB.CLASSIC. The runtime makes API calls and the value you get
from GetLastError may not be correct...

To get API error returns you should use Marshal.GetLast Win32Error().
And the calls to format message are not needed either. Just create an
instance of System.Componen tModel.Win32Exc eption and then look at the
.Message property...

' BitBlt code...
If intRetun = 0 Then
Dim ex As New Win32Exception( Marshal.GetLast Win32Error())
MessageBox.Show (ex.Message)
End If

--
Tom Shelton [MVP]
Powered By Gentoo Linux 1.4
Of all men's miseries, the bitterest is this:
to know so much and have control over nothing.
-- Herodotus

[VB Programmer]

Thanks everyone I will try it today.

"Armin Zingler" <az*******@free net.de> wrote in message
news:us******** ******@TK2MSFTN GP12.phx.gbl...

"Ken Tucker [MVP]" <vb***@bellsout h.net> schrieb

Are you getting an error ?

Untested: I think we could reproduce the problem by pressing Win+L and
having the app make a snapshot meanwhile...
--
Armin