C# Discussion Boards - CodeProject

simple regular expression problem

15-Jun-08 12:58

willybt88 wrote:
to a running Winforms app (not in .NET)

This is a contradiction. WinForms is .NET. What language is your Win32 app then ?

willybt88 wrote:
I imagine that this is going to require ActiveX, can someone just confirm that this is the correct approach?

If you're going to send it to an app running on the client, ActiveX is your best bet, yes.

Christian Graus

Please read this if you don't understand the answer I've given you

"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )

Stephen Po15-Jun-08 11:55

Stephen Po

15-Jun-08 11:55

hi,
I'm trying to use a regular expression to find all data following two blank lines.
this means there should be two newline formatters preceeding all data I want to find. therefore I've been trying something like:

(?<=\n\n).*

somehow I need to check the lines are indeed blank with only newline formatters and then return the text following the two new lines.
no luck though. any ideas?
thanks,
Stephen

Re: simple regular expression problem

Christian Graus15-Jun-08 12:09

Problem to Send Message to other Clients via MSMQ [modified]

15-Jun-08 12:09

Well, it's \n\r ( or \r\n, I forget, I use Environment.NewLine instead ).

hdv21215-Jun-08 11:15

hdv212

15-Jun-08 11:15

Hi i want to send message to other clients via MSMQ by this path :
[clientName]\Private$\Orders
but the followinng error has shown me :
Invalid queue path name
but if i use . (dot) as follow :
.\Private$\Orders
my queue was successfully created, where does my problem and how to solve it ?
i'm also get the following error when use this path :
FormatName:Direct=OS:[myMachineName]\private$\remote

error :
Can not create a queu with the path FormatName:Direct=OS:[myMachineName]\private$\remote
thanks

modified on Sunday, June 15, 2008 5:50 PM

GraphicsPath remove or alter line

Chris Copeland15-Jun-08 11:13

Chris Copeland

15-Jun-08 11:13

Hey.

I've been trying to get this to work but with no avail. I'm trying to modify a GraphicsPath to remove a line or move it so it matches the situation i'm using.

I'm using it as part of a new Rendering mode (for the ToolStripManager.Renderer), and i've got a code that creates a path (and adds curves to the corners if I want it to).

My code gets the GraphicsPath of the AffectedArea of a MenuStrip dropdown (or ToolStripDropDownButton). I want the code to remove or alter the line to leave a space for the ConnectedArea section.

Here's a picture of the area, I mean, that I want to be able to remove (opens in new window)

The black border is the GraphicsPath lines that the code retrieves. The red part is the part of the line I want to remove.

Here's a sample of the code inside OnRenderToolStripBorder(ToolStripRenderEventArgs e)

if (e.ToolStrip is ContextMenuStrip || e.ToolStrip is ToolStripDropDownMenu)
{
	Rectangle Area = new Rectangle(0, 0, e.AffectedBounds.Width - 1, e.AffectedBounds.Height - 1);

	e.Graphics.SmoothingMode = SmoothingMode.AntiAlias;

	using (GraphicsPath Path = CreateDrawingPath(Area, 0))
	{
		e.Graphics.DrawPath(Pens.Black, Path);
	}
}

I've tried using:

Path.PathPoints[0] = new PointF(e.ConnectedArea.Width, 0);

To try and change the first point of the GraphicsPath to start from the X axis of the width of the ConnectedArea (the GraphicsPath goes "Left-top" -> "Right-top", "Right-top" -> "Right-bottom" etc).

But it doesn't work. Does anyone have an idea how I might be able to move the first point to where the ConnectedArea Width ends?

Thanks in advance Smile | :)

Converting a project to .Net 3

Muammar©15-Jun-08 9:28

15-Jun-08 9:28

Hi,
I have an application basically written in VS 2003, then I upgraded to 2005, now I'm running 2008 and I cant use any of the .Net 3 components so please tell me how to upgrade my app to be .Net 3 compatible

Thank you guys!

All generalizations are wrong, including this one!
(\ /)
(O.o)
(><)

Christian Graus15-Jun-08 9:32

15-Jun-08 9:32

I don't see why you can't, there's no such thing as a .NET 2.5 project, your project is .NET 3.5 ( not 3.0 ) or it's not.

Muammar©15-Jun-08 9:37

15-Jun-08 9:37

Hi Christian,
It's good to see you Smile | :)

, I always get optimistic when the answer is from you Wink | ;)

You see, I had to use the element host thing to include some wpf controls but i just cant reference any of the .net 3 dlls when it's not the case when i create a new app in .net 3/3.5

i'm sure i just need to find a way to convert my whole app to support .net 3/3.5 but i dont know how to do it, there should be some tool somewhere withing the ide i suppose

All generalizations are wrong, including this one!
(\ /)
(O.o)
(><)

Christian Graus15-Jun-08 9:43

15-Jun-08 9:43

It seems probable to me that your project did not include references to the .NET 3 dlls, and you just need to add those references.

Muammar©15-Jun-08 9:49

15-Jun-08 9:49

Christian Graus wrote:
..references to the .NET 3 dlls, and you just need to add those references.

I thought so too at first but it's not possible when the project is not based on .Net 3 framework from the moment it is created!

You just cant include any dll with the version 3.x.x as it will be disabled!
Please try now to see what I mean, I'm sure you will be surprised as I am now Smile | :)

Create a new solution based on a version less than 3 and try to reference anything from .Net 3.

All generalizations are wrong, including this one!
(\ /)
(O.o)
(><)

Christian Graus15-Jun-08 9:52

15-Jun-08 9:52

Oh - there's a setting in 2008 that sets what framework you're building against, it DOES support 2.0. Perhaps the conversion is set to that by default, and you need to change that in the project settings ?

Muammar©15-Jun-08 9:53

15-Jun-08 9:53

Ok sorry

I got it

Properties > Target Framework: .Net 3.5
That was stupid Blush | :O

Thanks Christian!

All generalizations are wrong, including this one!
(\ /)
(O.o)
(><)

Christian Graus15-Jun-08 9:54

15-Jun-08 9:54

Glad to help Smile | :)