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Did you override OnPaint to do the drawing yourself? Did you call SetStyle(ControlStyles.AllPaintingInWmPaint | ControlStyles.OptimizedDoubleBuffer, true) in the form's constructor?
That will work if you do it correctly. Worse comes to worse, just draw to a bitmap, then draw that bitmap to the screen, which will prevent flickering.
Tech, life, family, faith: Give me a visit.
I'm currently blogging about: Little House on the Flickr
Judah Himango
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thanks. I try to do it again, and i find a mistake. If you dont say this. May be i dont find. Thanksss again...
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I have RichTextBox that i want to add an animated picture within it and i can be able to insert that picture wherever in a string as the that of yahoo messenger during private or public chatting i succeeded to insert it but non animated even if the picture is animated one but also the picture can be resized with the mouse of the user when he hold it the user can resize it but i don't want it to be resized and i want it to be animated if it is for example .Gif Picture. and to have the ability to insert it in any place within the string entered
How can i do that ?
Thanks everybody - "Human Knowledge belongs to the world"
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That was the longest run-on sentence I've ever seen; it caused me a good deal of pain to read that. Your English could use some help; but I'll assume you're not a native English speaker so I'll try to answer your question anyways.
The RichTextBox probably won't do everything you want it to do out-of-the-box; you're going to have to dig into some Win32 interop to get it to do things differently. Have you read the CP article[^] on customizing the RTB? There's another article on it here[^] that may help you as well.
Tech, life, family, faith: Give me a visit.
I'm currently blogging about: Little House on the Flickr
Judah Himango
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How Can i make the TreeView Nodes start from the right edge of the control with C# ?
M.Gamal
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myTreeView.RightToLeft = System.Windows.Forms.RightToLeft.Yes;
myTreeView.RightToLeftLayout = true;
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"TreeView dosnt contain RightToLeft RightToLeftLayout"
Iam using C# ,and I get this Error when using MyTreeView.RightToLeftLayout = true;
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Hi
I hope someone in here can help me with this issue.
I am trying to write a little application that will give me the information of the directory which is there in the network.
I am able to get the local directory information, but when I tried to access the information of the directory which is in some other
system of the same network I am getting Exception as "Logon failure: unknown user name or bad password."
In fact the directory which I am trying to access is created by myself using another console application.
Here is my code snippet,
DirectoryInfo dir = null;
FileInfo[] xmlFiles;
//get the directory info,
//here test machine is a network machine under same domain
dir = new DirectoryInfo("\\\\TestMachine\\Indexes\\");
//get the information about the subdirectories of dir.
DirectoryInfo[] subDirectories = dir.GetDirectories();
when I try to execute this GetDirectories() method its says "Logon failure: unknown user name or bad password."
So, am doing anything wrong in this code? or is there any other solution to get the directory Info?
Expecting some solution.
Thanks in advance.
Hans.
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The Genghis Framework[^], which is free and built on top of the .NET framework, includes functionality for accessing shares and password protected shares.
Also, you can use WNetUseConnection[^] from the PlatformSDK via P/Invoke.
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Basically, the remote machine doesn't trust you. You're trying to get the directory of a share on a machine that is not part of the same domain as the machine your code is running on. The SID that you logged in under on your machine isn't known to the remote machine and there is no way for it to know that it can be trusted, so it denies your "implied" login attempt.
You have to establish an authenticated connection to the remote machine before you attempt any file operations on one of its shares. You can use the library Judah suggested to establish the authenticated connection, but you still have to supply it with a username and password that the remote machine will accept as valid.
RageInTheMachine9532
"...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome
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Hai friends,
i have to export a table from any of the database types(Text,Access or Excel) to excel...provided that the datatypes of the Columns should retain...
To be clear....
I have to craete a new Excel Sheet whose datas will be imported from other database tables..along with their corresponding datatypes...
I tried to create a new excel importing datas from a textfile..but all datas are in general datatype format....How can i get the solution...
Kindly help me...
Thanx in advance...
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Hi,
i've tried 'google'ing to find any article on how i might be able to read the metadata of a file or .net 2.0 rc1 assembly that i've built (e.g. the date the file was last modified, or the author of the assembly, it's version etc), but i seem to have hit a bit of a brick wall, i'm sure there must be a simple (or not) way to read this kind of information about a file programmtically.
any help or nudges in the right direction will be greatly appreciated
Thanks in advance.
Afzal "AV-E" Hassen
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I'm not sure if it will give you all the info you need, but you can do:
Assembly asm = Assembly.Load(filename);
then use properties/methods of asm:
asm.GetName().Version
You can also get attributes of an assembly by using GetCustomAttributes(typeof(AssemblyTitleAttribute), true)
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Try this...
using System;
using System.IO;
using System.Reflection;
public class Meta
{
public static int Main( )
{
// First load the assembly.
Assembly a = Assembly.LoadFrom("hello.exe");
// Get all the modules that the assembly supports.
Module[] m = a.GetModules( );
// Get all the types in the first module.
Type[] types = m[0].GetTypes( );
// Inspect the first type.
Type type = types[0];
Console.WriteLine("Type [{0}] has these methods:", type.Name);
// Inspect the methods supported by this type.
MethodInfo[] mInfo = type.GetMethods( );
foreach ( MethodInfo mi in mInfo )
{
Console.WriteLine(" {0}", mi);
}
return 0;
}
}
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Thanks guys. i'll give it a shot
Afzal "AV-E" Hassen
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Hi gurus,
I'd like to know how to do the two following operations on TextBox controls:
- Moving the cursor to the end of the text, so the new added lines are always visible without moving the scroll bar.
- Setting the focus on a TextBox. the Focus() method does not work 
Thanks for the help.
fred.
There is no spoon.
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answers:
1. <textboxname>.Select();
SendKeys.Send("{END}");
2. <textboxname>.Select();
Sheel Gohe
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1) I have found another alternative by using the SelectionStart set to Text.Length and ScrollToCaret method.
2) The Select() method does not set the caret to the wanted control. Actually the Focus() method was right, the wrong thing was that I was modifying the interface through a secondary thread.
Thanks for the help.
There is no spoon.
-- modified at 11:05 Wednesday 18th January, 2006
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this is my class code
i wan't to get ip address of each client which is getting connected
listen class
public class listen
{
TcpListener server=null;
Thread tcpthread=null;
client[] cl=new client[3];
public listen()
{
//
// TODO: Add constructor logic here
//
}
public void startlisten()
{
Int32 port = 3310;
IPAddress localAddr = IPAddress.Parse("192.168.0.6");
// TcpListener server = new TcpListener(port);
server = new TcpListener(localAddr, port);
// Start listening for client requests.
server.Start();
// Enter the listening loop.
for(int i=0;i<3;i++)
{
cl[i]=new client();
cl[i].status=true;
}
Boolean flag;
while(true)
{
try
{
flag=false;
// Perform a blocking call to accept requests.
// You could also user server.AcceptSocket() here.
for(int i=0;i<3;i++)
{
if(cl[i].status==true)
{
cl[i]= new client(serverAcceptTcpClient());
tcpthread=new Thread(new ThreadStart(cl[i].getClient));
tcpthread.Start();
flag=true;
break;
}
}
if(flag!=true)
{
//MessageBox.Show("All Clients Busy");
}
}
catch(Exception se)
{
}
}
}
public void stoplisten()
{
server.Stop();
}
}
client class
public class client
{
TcpClient tcpClient;
public Boolean status;
// Buffer for reading data
Byte[] bytes = new Byte[256];
String data = null;
IPAddress localAddr = IPAddress.Parse("192.168.0.6");
public client()
{ //
// TODO: Add constructor logic here
//
//status=true;
}
public client(TcpClient Client)
{
tcpClient =Client;
//
// TODO: Add constructor logic here
//
status=false;
}
public void getClient()
{
try
{
data = null;
// Get a stream object for reading and writing
NetworkStream stream = tcpClient.GetStream();
int i;
// Loop to receive all the data sent by the client.
while((i = stream.Read(bytes, 0, bytes.Length))!=0)
{
// Translate data bytes to a ASCII string.
data = System.Text.Encoding.ASCII.GetString(bytes, 0, i);
// Process the data sent by the client.
string replyMsg = data;
clamdCommand x=new clamdCommand();
replyMsg=x.Command(replyMsg);
int lowerport=int.Parse(ConfigurationSettings.AppSettings.Get("lowerport"));
int upperport=int.Parse(ConfigurationSettings.AppSettings.Get("upperport"));
for(int y=lowerport;y
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You forgot to mention what your problem is.
---
b { font-weight: normal; }
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i want to get ip address of each client on server when it is getting connected.
Above code is for server
how can i get "RemoteEndPoint"
but not getting
Ankit Aneja
"Nothing is impossible. The word itself says - I M possible"
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But what is the problem? Do you get an error message?
---
b { font-weight: normal; }
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Hello
i have wasted lot of time.
1I want to put a command button in the datagrid in one of the column. I got a piece of code from this site and i have added to my project. it is working fine except i am having a major issue. when i click on the button, i am poping a form. if i click again, the form gets popped up twice, if i click again, 3 times the form gets popped up. is it due to the events get added up every time when i click on the button. my code is like this
buttonColStyle = new DataGridButtonColumn(1)
buttonColStyle.HeaderText = "Browse";
buttonColStyle.MappingName = ObjDataColumn.ColumnName;
buttonColStyle.CellButtonClicked +=
new DataGridCellButtonClickEventHandler(HandleCellButtonClick);
GridTableStyle.GridColumnStyles.Add(buttonColStyle);
this.DGDetails.MouseDown += new MouseEventHandler(buttonColStyle.HandleMouseDown);
this.DGDetails.MouseUp += new MouseEventHandler(buttonColStyle.HandleMouseUp);
wilson
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dear wilson you have to check the instance of the popped-up form every time you click the button. If form's instance already exist then dont make use of new keyword just use the form's "show" method.
Use this method to check instance of the form (may be this could be useful)
//***********Start**********
private Form isFormAlreadyLoaded(Form ToBeLoadedForm)
{
foreach (Form LoadedForm in this.MdiChildren)
{
if ((LoadedForm.Name == ToBeLoadedForm.Name) && ((LoadedForm.Text.Trim ()) == ToBeLoadedForm.Text.Trim()))
{
return LoadedForm;
}
else
{
return ToBeLoadedForm;
}
}
return ToBeLoadedForm;
}
//***********End**********
regards
Sheel Gohe
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I have a Windows application written c# 1.1 that uses embedded crystal reports. I have 60 users who all use the exe which is accssed from a shared folder on the network. The problem is I still need to install the application on all client machines so as crystal registers properly. Is there away around this so as I dont have to install the application on all client machines?
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