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Greetings!
I want to create project with video abilities.
Here is short description of my needs:
Application needs to be done with .NET Framework.
Application should consist from two parts:
Client application and server application have different functionality, but they both should work with web-cams installed on their sides (client and server sides).
First part is a client application.
Client application can connect to different web – cams and see content of web – cam.
User of client application can talk to server application using microphone and listens answer from server user.
Except this, client application should have ability to save video from server side.
Server application should have similar video abilities with client application.
I analyzed different technologies for my purpose and I lost one's the head in technologies.
Please suggest me what technology best suites for my project.
Thank you all for your attention for my question
Best Regards,
Alex
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Hi
I have a problem with set cursor into TextBoxes, which are usually unvisible.
After some events the TextBoxes I set some TextBoxes to visible. To set the cursor to the actually visible TextBox use Select() or Focus(), but the cursor doesn't change it's position and stay in the last TextBox, only after a minimize and maximaize action or change to the other programs and back to this program moved to the expected TextBox. Invalidate or Refresh and DoEvents cann't help. I think it's something wrong with exessive using of my internal events, because in a testprogram witout my events the cursor behave as I expected.
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Hi
after insertion some lines the vertical scroll is coming, but the RichTextBox still show the upper side and doesn't scroll to the last line and doesn't show the last inserted line. Should I get the count of lines and scroll line by line?
Info: This RichTextBox has't any focus. It's only a logging MultiLine-TextBox .
BR
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I had the same problem with a listbox: I do not know a way to have it scroll automatically
to keep the bottom line visible, so this is what I came up with (and I assume something
similar applies to a RichTextBox):
public void Output(string s) {
lb.Items.Add(s);
if (Control.MouseButtons==MouseButtons.None) {
lb.TopIndex=lb.Items.Count-1;
}
}
BTW: the value for the top index is not correct
(it should actually be Items.Count - numberOfLinesVisibleInListbox)
but it works just fine as is, so why bother. (But the -1 is necessary).
Luc Pattyn
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I'm sorry if I post this here, because I cannot find the thread about Compact Framework. Well, to the point. We know that we can use custom image as the toolbox item (Framework). We just use the attribute ToolboxBitmap to set the component/control icon. But at .NET Compact Framework 2.0 I cannot find the attribute like this, does anyone know alternate way to use custom icon for the component/control (on designer for sure not runtime).
Thanks for helping.
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It looks likes it's not supported. BTW: The CF Forum is called "Mobile Development".
The alternative is to add a 16x16 pixel BMP as an embedded resource with the same name as the control class.
Dave Kreskowiak
Microsoft MVP - Visual Basic
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With namespace or not, e.g class name is "classA" in namespace "nsA", so on resource use "nsA.classA.bmp" or only "classA.bmp"?
Thanks for the reply.
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Shouldn't need it. If should just be classname.BMP.
Dave Kreskowiak
Microsoft MVP - Visual Basic
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I tried that but not working, I'm VB.net, my component is in this namespace "CRMSoft.Controls.Forms.NumberPicker.vb" Bitmap and left it at "CRMSoft.Controls.NumberPicker.bmp" not right?
Cleber Roberto Movio
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Hi
I would like to list all the available servers from tnsnames.ora file.
DBNAME =
(DESCRIPTIOMN= (ADDRESS_LIST=(ADDRESS=(PROTOCOL=TCP)(HOST=113.114.2.145)(PORT=1521))
(CONNECT_DATA=SERVICE_NAME=dbname))
In this file i want to read exactly "service_name = dbname". How to do it in C#. Any one could you help me please?
kesavan
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Please don't cross post.
the last thing I want to see is some pasty-faced geek with skin so pale that it's almost translucent trying to bump parts with a partner - John Simmons / outlaw programmer
Deja View - the feeling that you've seen this post before.
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Ok another problem with .NET Remoting. 
Server create object in mode='singleton' by file configuration. Object has member 'parent_'.
Like this:
-----------------------------------------------------------------------
RemotingConfiguration.Configure(@"TestServer.exe.config", false); remoteobject_ = new RemoteObject();
remoteobject_.parent_=this
-----------------------------------------------------------------------
I tried connect to this object in client like this:
-----------------------------------------------------------------------
RemotingConfiguration.Configure(@"TestClient.exe.config", false);
WellKnownClientTypeEntry[] ClientEntries = RemotingConfiguration.GetRegisteredWellKnownClientTypes();
remoteobject_ = (RemoteObject)Activator.GetObject(typeof(TestRemothing.RemoteObject),ClientEntries[0].ObjectUrl);
-----------------------------------------------------------------------
and i noticed, that client work with other remote object than server. Why? How client can work with object, that already created by server?
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 I fined solution! Thank's codeproject 
qqq
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By using this, what exactly is locked? A specifiek variable or the thread itself or the variables that have been modified with 'System.Threading.Interlocked.Increment'?
I THINK 'THE VARIABLES THAT HAVE BEEN MODIFIED BY THE THREAD'. WRIGHT?
If you have the following:
public static void Synchronization_WriterLock()
{
System.Threading.ReaderWriterLock oReaderWriterLock;
oReaderWriterLock = new System.Threading.ReaderWriterLock();
miCount = 0;
try
{
oReaderWriterLock.AcquireWriterLock(100);
try
{
System.Threading.Interlocked.Increment(ref miCount);
Console.WriteLine(miCount);
}
finally
{
oReaderWriterLock.ReleaseWriterLock();
}
}
catch (ApplicationException e)
{
Console.WriteLine("Failed to get a Writer Lock");
}
}
What locks what?
-- modified at 17:03 Wednesday 27th December, 2006
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Can anybody help me  . I use remote object with events. When client connect to event - system throw SecurityException:
"Message = "Type System.DelegateSerializationHolder and the types derived from it (such as System.DelegateSerializationHolder) are not permitted to be deserialized at this security level."
In MSDN i finded this:
--------------------------------------------
To adjust the security level
1. Modify the configuration file(s) generated in conversion.
- or -
2. Edit the converted code.
--------------------------------------------
but how to do it?
qqq
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OK. I find solution! Thanks 
qqq
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Hi all
I am developing one installation wizard. I have one issue. If i click the next button I made the current form visible = false and creating instance for the next from and showing the form. There problem I facing is, I dont know how to show the previous form by hidding the current form. Can anybody help me please....
Thanks
kesavan
kesavan
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CurrentFormName.Hide
PreviousFormName.Show
Obviously that's VB code, but you should be able to figure out the C# equivalent pretty easily.
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Just to be clear - if you're inside the current form, you need to only call Hide, and specifying the name of the form classes won't work, you need to specify the name of form instances that are visible within the scope of the method that's got the call in it
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
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I was expecting the result of this search for "1" to be at location "0"
<br />
class Program<br />
{<br />
static void Main(string[] args)<br />
{<br />
List<double> listItems = new List<double>();<br />
<br />
listItems.Add(1);<br />
listItems.Add(2);<br />
listItems.Add(4);<br />
listItems.Add(6);<br />
listItems.Add(4);<br />
listItems.Add(3);<br />
listItems.Add(8);<br />
listItems.Add(4);<br />
listItems.Add(1);<br />
<br />
listItems.Sort();<br />
<br />
for (int i = 0; i < listItems.Count; i++)<br />
{<br />
Console.WriteLine("{0} : {1}", i, listItems[i]);<br />
}<br />
<br />
Console.WriteLine(listItems.BinarySearch(1));<br />
}<br />
}<br />
However, the result is at location "1". Any reason for this?
With love,
Paul.
Jesus Christ is LOVE! Please tell somebody.
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Paul
In your list, you have two entries for 1. You sort your list, so both those entries will move together, i.e. be at position 0 and 1. Now, a binary search works by starting at the center of the list, and then deciding whether or not the value has been found or is higher or lower. In this case, it is lower. It effectively discards the top of the list and moves to the center of the lower selection and repeats this process again. It keeps going until it finds a match. In this case, it will find the item at position 1 because that is the first match it will find. I hope that this makes sense to you.
the last thing I want to see is some pasty-faced geek with skin so pale that it's almost translucent trying to bump parts with a partner - John Simmons / outlaw programmer
Deja View - the feeling that you've seen this post before.
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Pete O`Hanlon wrote: I hope that this makes sense to you.
It does. That it starts from the center, I know but I think as an algorithm, there are checks to account for cases like this.
I have tested a case of three "1", and the result is position "1", and for four "1", the result is position "2".
So, may be the question should be should conditions must be there before we consider binary search?
With love,
Paul.
Jesus Christ is LOVE! Please tell somebody.
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Your array after sorting would look like 1 1 2 3 4 4 4 6 8
If you look at the way binary search works, the first condition of it would be
low = 9
high = 0
mid = 4
if (collection[4] = SearchItem) -> false
now since the search item wasnt found and the search item is less than the mid value, the elements that would be used to search next would be 1 1 2 3
low = 3
high = 0
mid = 1
if (collection[1] == SearchItem) -> True, so it returns the position of mid which is 1
Tarakeshwar Reddy
MCP, CCIE Q(R&S)
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I have tested a case of three "1", and the result is position "1", and for four "1", the result is position "2".
So, may be the question should be should conditions must be there before we consider binary search?
With love,
Paul.
Jesus Christ is LOVE! Please tell somebody.
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