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Yep, you are right... but we want to confuse beginners, or? 
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sir Actually i am use dotnet after 1 year thats why i am facing lot of problem
sir i kindly request to u please give me solution 
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Try the below sample code
query = "select username from tablename where username = '" + usr.text + "'";
sqlcommand cmd = new sqlcommand(query,con);
sqldatareader dr;
dr=cmd.executereader();
while(dr.read())
{
if(dr["username"].tostring() == usr.text)
{
true user
}
else
{
false user
}
}
dr.close();
Regards,
Karthik K...
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Be careful about how you test for equality, in all but the rarest case you shoud use the Equals method.
a == b
This compares two objects and returns true if they are they same object, that is the same instance. This does not compare the content.
a.Equals(b)
Compares the two instances and is defined by each classes implimentation. In the trivial case of the string, it returns true if they have the same content. Any class you write should override this method if you need to compare instances' values.
Panic, Chaos, Destruction.
My work here is done.
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My WCF properly working when SecurityMode.None. If I switch to SecurityMode.Message so it is not work in client side. My main problem is how can I encrypt to message in WCF?
Following code OK, its work;
Host Side Code
---------------------------------------------------------------------
NetTcpBinding tcpBinding = new NetTcpBinding(SecurityMode.None);
Uri tcpUri = new Uri("net.tcp://10.0.105.240:7070/HesaplamaServisi");
ServiceHost host = new ServiceHost(typeof(Hesaplayici), tcpUri);
host.Description.Behaviors.Add(new ServiceMetadataBehavior());
host.AddServiceEndpoint(typeof(IHesaplamalar), tcpBinding, "net.tcp://10.0.105.240:7070/HesaplamaServisi");
host.AddServiceEndpoint(typeof(IMetadataExchange), System.ServiceModel.Description.MetadataExchangeBindings.CreateMexTcpBinding(), "mex");
host.Open();
Client Side Code
---------------------------------------------------------------------
NetTcpBinding tcpBinding = new NetTcpBinding(SecurityMode.None);
IHesaplamalar proxy = ChannelFactory<IHesaplamalar>.CreateChannel(tcpBinding, new EndpointAddress("net.tcp://10.0.105.240:7070/HesaplamaServisi/"));
FOLLOWING NOT WORK
---------------------------------------------------------------------
I want encrypt to message, so I add following code both side;
NetTcpBinding tcpBinding = new NetTcpBinding(SecurityMode.Message);
tcpBinding.Security.Message.AlgorithmSuite = System.ServiceModel.Security.SecurityAlgorithmSuite.TripleDesSha256Rsa15;
This code properly work when Host and Client same PC, If client.exe run another PC in same network this code not work. What can I do? Thanks...
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Hi,
it is just a guess but which operating systems are you using for client and server? As far as I know is SHA256 only supported on Vista or higher. Maybe thats a problem.
Regards
Sebastian
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By the way, did you try to debug your code? Maybe you get an exception for this.
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I made the following class:
public class MyTextBox : TextBox
{
private int _IntVariable = int.MaxValue;
private myEnum _EnumVariable;
public MyTextBox()
{
}
public int IntVariable
{
get {return _IntVariable;}
set {_IntVariable = value;}
}
public myEnum EnumVariable
{
get {return _EnumVariable;}
set {_EnumVariable = value;}
}
}
When I create a user control and place a "MyTextBox" control on the user control, I name it txtName and set the IntVariable to 7 and the designer shows 7. Then, when I build the solution again, the value shown in the designer resets to the maximum value of an integer while the code in the InitializeComponent() method still shows : this.txtName.IntVariable = 7;.
However, the enumeration property can be set and it persists when the solution is re-built, the UserControl is closed and re-opened in the designer, or Visual Studio is closed and re-opened and the solution re-opened.
Can anyone explain why the integer property is not persisting and the enumeration property is and if there is anything I can do to make the integer property persist in the designer?
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Sorry, I can't replicate the problem. It behaves as expected here 
DaveBTW, in software, hope and pray is not a viable strategy. (Luc Pattyn)
Visual Basic is not used by normal people so we're not covering it here. (Uncyclopedia)
Why are you using VB6? Do you hate yourself? (Christian Graus)
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Hi,
I'm trying to build an over laying image like in the following link if you do mouse over Stars after Rate this Content you will see the overlaying form.
http://msdn.microsoft.com/hi-in/library/a72418yk(en-us).aspx[^]
I want to build same kind of control in C# Windows application? Is it possible ? Please help out.
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IMO PictureBox which displays a Panel with needed controls on MouseHover might be a good start.
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How to open a new File on a Button click such as an Application file.
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Have you tried to Google[^]for it??
Excellence is doing ordinary things extraordinarily well.
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Use an OpenFileDialog, then open the file. How you do this depends on which file format it is. If you just want to open it, use Process.Start(fileName)
Between the idea
And the reality
Between the motion
And the act
Falls the Shadow
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Rupaa wrote: How to open a new File on a Button click such as an Application file.
To do anything on a button click you will require event processing.
If you want to open a file specified by the user, so you need only a file path in string format, you could use OpenFileDialog.
If you want to open a file, so actually access it, which would mean you want its information. You can use
some .Net's classes inside of the System.IO namespace. They are ending with 'reader'. So what type of file access do you need?
If you want just only to execute a certain assembly executable, then I guess Process.Start(fileName) should be fine.
... I hope this helps! 
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can you help me to build an application use scanner to scan document
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Yes I can.
My rates are $120 (US) per hour.
Let me know
___________________________________________
.\\axxx
(That's an 'M')
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_Maxxx_ wrote: My rates are $120 (US) per hour
Mine $119.50 (US) per hour
himanshu
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Too expensive. I would do at $119. I can see you losing business. 
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How about 120 rupees?? 
Excellence is doing ordinary things extraordinarily well.
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If that's per second, I can manage.
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For that you need to use the Scanner SDK to create your own scanning application. There are lots of 3rd party scannner SDKs available for you.
But if you need a simple application then this may help. .NET TWAIN image scanner[^]
Books are as useful to a stupid person as a mirror is useful to a blind person. - Chanakya
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So I am trying to make a class that uses a constructor of a different class.
I use the code (Name of class) Class = new (Name of class(constructor variable))
But oddly the constructor variable won't appear. Could anyone please help?
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Your question is not clear. Posting relevant code will help.
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public class CharClasses
{
public int healthPoints = 0;
public int manaPoints = 0;
public int strength = 0;
public int dexterity = 0;
public int speed = 0;
public int intelligence = 0;
public CharClasses (string characterClass)
{
if (characterClass == "Warrior")
{
healthPoints = 100;
manaPoints = 10;
strength = 10;
dexterity = 5;
speed = 3;
intelligence = 0;
}
if (characterClass == "Archer")
{
healthPoints = 75;
manaPoints = 10;
strength = 3;
dexterity = 10;
speed = 5;
intelligence = 0;
}
if (characterClass == "Assassin")
{
healthPoints = 75;
manaPoints = 10;
strength = 5;
dexterity = 3;
speed = 10;
intelligence = 0;
}
if (characterClass == "Mage")
{
healthPoints = 50;
manaPoints = 50;
strength = 0;
dexterity = 5;
speed = 3;
intelligence = 10;
}
}
}
public class Leveling
{
public int level = 1;
public int experience = 0;
public Leveling()
{
if (experience == ((experience + 100) * 1.5))
{
level++;
}
}
}
I'm trying to take the constructor named "characterClass" from the class CharClasses and use it in the class Leveling.
CharClasses Class = new CharClasses(characterClass); I'm trying to make this code work but the characterClass part of it does not seem to work...
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