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Hi,
I have a problem with a datagrid.
I want to add an extra row to it, this works fine by building tablecells and add them to the grid for each entry in the grid.
The extra rows can be edited by the user, but I don't know how I can bind this extra rows to my datagrid, so the changes of the user are recognized by the webform. 
can you please help me with my Problem, or is it impossible to add extra rows, that can be edited by a user?
thank you for your help
Peggy
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You do not need to do anything extra. You create a datatable and bind it to datagrid. Allow user to add a row to datagrid. When user says update then you just use SqlCommanBuilder object to perform the insert operation. You can also refer my article http://www.codeproject.com/useritems/DataGridView_manipulation.asp[^] 
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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Hello Friends,
I want to know how to use assembly which is registered in GAC(Global Assembly Cache).
I have Created an assembly name dellMe.dll and i have registered it with same name but different
version 1.0.0.0 abd 1.0.0.1. Can i load the any one of the assembly at runtime.
Basically, i want to know how to do this refering of assembly in our C#.Net application.
Any Example will be helpfull.
Thanks In Advance.
Rahul Kulkarni
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You can do it using reflection.
Assembly asm = Assembly.Load("System.Data, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089");
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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For this i have to copy the assembly. Right?
how can i keep different versions of assembly of same name?
Thanks.
Rahul Kulkarni
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If you check the code carefully then you will come to know that System.Data is from GAC only. Now I have .NET 1.1 and .NET 2.0 installed on my machine. Both contains System.Data but with different versions. That is the speciality of GAC. You need to create a strong name and change the version. Then you can copy your assembly in GAC. You do not need to copy it to your application.
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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Thanks ....It works nice.
Thank You Very much
Rahul Kulkarni
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As per you said all working fine.
even we can create instatance of the class of the assembly.
How can we call function of that class. It is not giving intellisense.
what is the solution.?
Rahul Kulkarni.
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Can do like this.
Assembly asm = Assembly.Load("System.Data, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089");
Type[] tps = asm.GetTypes();
foreach (Type tp in tps)
{
if (tp.IsClass)
{
MethodInfo[] methods = tp.GetMethods();
object abc = Activator.CreateInstance(tp);
if (!methods[0].IsStatic)
{
methods[0].Invoke(tp, new object[] { });
}
}
}
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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Here is a sample code that will open a connection to sql server using reflection.
string connString = "Server = localhost; Database = EmployeePortal; User ID = abc; Password = abc; Trusted_Connection = False;";
Assembly asm = Assembly.Load("System.Data, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089");
Type[] tps = asm.GetTypes();
foreach (Type tp in tps)
{
if (tp.IsClass)
{
if (tp.Name == "SqlConnection")
{
Console.WriteLine("Sql connection found...");
}
MethodInfo[] methods = tp.GetMethods();
object abc = null;
try
{
abc = Activator.CreateInstance(tp, connString);
}
catch (Exception ex)
{
}
foreach (MethodInfo method in methods)
{
if (tp.Name == "SqlConnection" && method.Name == "Open")
{
Console.WriteLine(method.Name);
method.Invoke(abc, null);
}
}
}
}
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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Firstly, Thanks.
You helped me alot.
I got the concept you are telling.
On the basis of your example, i tried for messagebox function.
But it is giving me exception "Parameter Missmatch".
I am not getting were do i can function returned value
Rahul Kulkarni
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If you have overloaded methods then you need to check which method you want to execute. Here is the sample code to call MessageBox.Show("Hi").
Assembly winAsm = Assembly.Load("System.Windows.Forms, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089");
Type[] allTyps = winAsm.GetTypes();
foreach (Type typ in allTyps)
{
if (typ.Name == "MessageBox")
{
Console.WriteLine("MessageBox found...");
MethodInfo[] mtds = typ.GetMethods();
foreach (MethodInfo mtd in mtds)
{
if (mtd.Name == "Show")
{
if (mtd.GetParameters().Length == 1)
{
mtd.Invoke(typ, new object[] { "Hi!!" });
}
}
}
}
}
Jayant D. Kulkarni
Brainbench Certified Software Engineer in C#, ASP.NET, .NET Framework and ADO.NET
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Hello! I am developing a GUI and I am having an strange problem. It
should be a very easy thing, but I have been walking around it for a
while. In my Form, I have a picturebox. I want to detect when the
mouse pointer enters the picturebox area WHILE the mouse left button
is pressed.
I have tried mouseenter, mousedown and mousemove events and none of
them are rised when I enter the control area with the mouse pointer
WHILE the left button is clicked!!!
Any ideas?
Thank you very much in advance!
Alvaro
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The event you want to treat is fired only if the none of the mouse buttons are pressed.
To overcome this behaviour, treat the MouseMove event of the form like this:
private void Form1_MouseMove(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Left)
{
Point pt = pictureBox1.PointToClient(Cursor.Position);
Rectangle rc = pictureBox1.ClientRectangle;
if (rc.Contains(pt))
{
}
}
}where pictureBox1 is your PictureBox object
SkyWalker
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Hi, Skywalker.
I have tried your solution, but this event is only fired when the mouse is moved in the form, outside the controls (I mean, move it around the background), so as soon as I move my mouse in a picturebox, it is not firing.
I am sorry... maybe trying to simplify the problem I haven't shown the correct enviroment. I will try it again:
I have a form with a panel, and in that panel I have 5 pictureboxes. When the user clicks (and hold the mouse button pressed as if he were dragging) into one of these five pictureboxes, I set a "origin" integer var to 1-5. What I want to do is, when the user moves the mouse with that button pressed and pass over one of the other picturbox, set a "target" integer variable with a value between 1 and 5.
The user could, in example set a origin and drag over several of the other pictureboxes updating the "target" var without release the mouse button.
Any ideas or comments would be very appreciated.
Thank you very much in advance!!!
Alvaro
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Believe me, I've tried the code before posting it
MouseMove is fired alright!
SkyWalker
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Clever idea!
got my 5
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Vielen Dank
SkyWalker
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What you are suggesting works in the case you click the left button in the main form, outside the pictureboxes. If you create two pictureboxes: picturebox1 and picturebox2 in your form and assign picturebox1 the mousedown event and the whole form the mousemove event you proposed,you will reproduce my problem. What I want is: clicking in area of the picturebox and holding the button pressed (drag, if you prefer), go to the area of the picturebox2 and obtain the message: "I have gone from 1 to 2" while my mouse button is still pressed.
I hope the code below clarify the issue.
int origin=0;
int target=0;
private void Form1_MouseMove(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Left)
{
Point pt = pictureBox2.PointToClient(Cursor.Position);
Rectangle rc = pictureBox2.ClientRectangle;
if (rc.Contains(pt))
{
target = 2;
// Do what you wanted to do within the MouseEnter event handler
MessageBox.Show("I have gone from " + origin + " to " + target);
}
}
}
private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
origin = 1;
}
Any ideas? Thank you for all your efforts!
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So what!
Copy the same code within the MouseMove event associated to the other control you want
( private void OtherPicture_MouseMove(object sender, MouseEventArgs e) )
And if you want a more general approach, then you have to do a little effort, but the idea is the same Try!
SkyWalker
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And it's still clever!
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OK. I have done what you have recommended, but it is unuseful. I have written then next mousemove event fot both the inkpictures:
private void inkPicture_MouseMove(object sender, MouseEventArgs e)
{
if (dragging)
{
PictureBox pb = (PictureBox)sender;
if (e.Button == MouseButtons.Left)
{
if (pb == pictureBox1)
{
target = 1;
}
else if (pb == pictureBox2)
{
Console.WriteLine("Asigned 2");
}
}
}
}
What I want is to get both the message "Asigned 1" and "Asigned 2" WITHOUT RELEASE THE MOUSE BUTTON, I mean, clicking and holding the mouse button in one of the two pictureboxes and going over the other one with the button still clicked. Please, try it. If you keep the button pressed when you leave the original picturebox, it remains throwing the mousemove event for that picturebox you have just leaved as if you were still in its area. That is my problem, and that is why I cannot detect when entering the new picturebox. I really hope my explanation to be clearly enough this time.
Any ideas?
Thank you very much in advance!!!
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Finally I got it. It seems that there is some kind of problem with mousemove when the mouse button is pressed and you get out of a control area, so finally I programmed it manually (this is what I was trying to avoid, Skywalker). I also had some problems with the "Contains" method, so I programmed it manually too. Here is the code for the mousemove event for both the pictureboxes:
public bool Contains(Rectangle r, Point p)
{
return (p.X >= r.X && p.Y >= r.Y && p.X <= (r.X + r.Width) && p.Y <= (r.Y + r.Height)) ? true : false;
}
private void inkPicture_MouseMove(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Left)
{
Point pt = this.PointToClient(System.Windows.Forms.Cursor.Position);
if (Contains(pictureBox1.Bounds, pt))
{
Console.WriteLine("Assigned 1");
}
else if (Contains(pictureBox1.Bounds, pt))
{
Console.WriteLine("Assigned 2");
}
}
}
}
Thank you ver much! 
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Hi
Pls help i'm looking alogorithm for encryption code to secure information using remote access. and how to connect Ms Access(Database) with java to store password and username in database.
zskhumalo
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