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What you're trying to do, makes no sense. What do you want to achieve ? Are you hoping to create a table in the DB ? Are you hoping to create a local table and use SQL on it ? You can specify a dataview, which operates SQL on a datatable, at least, I think you can. why don't you just create a table in your database ?
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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i ve to do it with dataview.but i want to know how i can add it to db n then display through dataadapter.
Chohan
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Well, wait a minute. You don't want to write code that adds tables at random, do you ?
you need to write the SQL to create the table, and execute it, through a connection, probably calling ExecuteNonQuery, as you want nothing back. Then you need to run some inserts, if you want some data to select from.
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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yeah.i want to add tables,insert columns ,rows and then display data of that tables.
Chohan
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Why would you want to write a program that keeps adding tables to your database ? What for ?
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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now for learning purpose
Chohan
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OK, what you've hopefully learned is, create your tables in SQL, then query them using SQL in your C# code.
Christian Graus - C++ MVP
'Why don't we jump on a fad that hasn't already been widely discredited ?' - Dilbert
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Hi friends,
In my C#.Net application i am trying to sort the gridview contents by using the following line.
dgAddressBook.Sort(dgAddressBook.Columns[23], ListSortDirection.Ascending);
but i am receiving the following error. Please help me to resolve this problem.
Error:
The original exception is: DataGridView control must be bound to an IBindingList object to be sorted. at System.Windows.Forms.DataGridView.SortDataBoundDataGridView_PerformCheck(DataGridViewColumn dataGridViewColumn)"
Thanks in Advance,
Regards,
Prya
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this is asp.net forum ....
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Hi all
I add usercontrol in code behide but it not run some event in (file ascx)
UserControl control = (UserControl)Page.LoadControl("tool/WebUserControl1.ascx");
Page.Controls.Add(control);
Can u show me how is postback to server on file ascx
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Hi,
I have this project it has a folder that it's supposed to read but when I run an error occurs the type or namespace name src could not be found(areyou missing a using directive or an assembly referrence?).
This folder is src. What could this be?
Thanks
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hi there:
is this compilation error not run-time?
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This error appears when I'm trying to run the project.
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can you post the code snippet here? anyway, the error message is rather clear that you missing the assembly...
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public src.controls.cont dngchart;
src is the folder that holds a folder controls wich holds the object cont. Do you think I can create a dll for it?
Thanks
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but in your solution do you make referecence to an assembly that cotain src.controls.cont namespace?
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What should I do?
Thanks.
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I'm not sure as I dont know what you really need... So what is the src.controls.cont? if that is a dll, just add reference to this...
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I create a crystal report in a asp.net web application. For printing i used window.print() javascript function. But the problem is i can't set page break. Co i used crystal report default printing method like MyReport.PrintToPrinter(1,False,0,0) method. Here it print with page breaks. But here also one problem I want to set printer name. for localhost it work perfectly. But in server how i can set printer name. It shows error while finding printer name. In some document i find one solution as map each and every printer in server. But how can it possible for an Internet application. Can u give one solution...
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in my project i want one panel control to be visible at on mouseover of an event i have writtten the code like this
in code view
lnkbtnwhoweare.Attributes.Add("onmouseover", "show('" & Me.pnlwhoweare.ID & "')")
and in html view
function show(id)
{
document.getElementById(id).style.visibility="true"
}
but it won't work what i have to do
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hi there:
try this document.getElementById(id).style.visibility = 'visible'...
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