In this example, it is not necessary since the example doesn't do anything much. The reason to do so would be if other classes implemented the interface and you want to call display() for a collection of different classes using a common variable of type Ia. GregD (My Blog on software development)

Do you have the 'Default Open Mode' set to 'shared' in the Database Options? Steve

Master Page would be your choice.

did you try the sort propert of DataView? Nav.

Are you sure that the project has reference to the System.Design.dll?

What you declare in code behind is only the reference to the control, not the control itself. --- b { font-weight: normal; }

I think that it is because you add controls using Controls.add() on its own. It may place the controls somewhere in the page that is not between form. Check that in the html generated through your browser. Try using a placeholder where to add your controls and you will have no problems at all cycling through it getting controls infos.

if you want to see the selected value each time, it is easy... give it an ID, say idListBox. while creating it, get the value from form variables to determine the selected value: ... selVal = request.forms("idListBox") loop create a list item if value of the item = selVal then add "selected" to the item add item end loop ... That's it... HainKurt

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insertAdjacentHTML Again, you can find much more info on this and other DOM related functions and properties by researching. The previous post was meant as an example to show the concept not an actual solution.

The frustrating thing is that you will have to use if statements in C# to do this....

hi, obviously yessssss.............. vivek 9868079257 delhi vivek

Hi sishya, 16,1 are savirity,status parameters for RAISERROR procedure.Don't think much about then you can get details on net regrading this. As your core problem To access this RAISEERROR return value....here is code. private void someMethod() { try { // Do a RAISERROR inside of a SQL block like // RAISERROR('You did something wrong with the value "%s"', 16, 1, @param1) } catch (SqlException sqle) { lblMessage.Text = sqle.Message; } catch (Exception err) { // Do something else with the exception } } I hope this will help you. Regards, Nagraj Let's Teach Life To Laugh........:laugh:

I am not sure if this is the best solution but just try it. Use javascript to display the name on mouse over,steps are as follows 1.try to add a div tag in html file: 2.add this in code behind - ddl is name of your drop down list ddl.Attributes.Add("onmouseover", "showTooltip();"); ddl.Attributes.Add("onmouseout", "showTooltip();"); i.e on mouseover you can see the name on mouse out the name disappears 3.Add the following javascript function showTooltip = function () { var obj = document.getElementById("ddl"); if(event.type == "mouseout") { document.getElementById("tooltip").style.display = "none"; } else { document.getElementById("tooltip").innerHTML =obj.options[obj.selectedIndex].text; document.getElementById("tooltip").style.display = "inline"; document.getElementById("tooltip").style.backgroundColor="#ffffcc"; document.getElementById("tooltip").style.paddingTop="1px"; document.getElementById("tooltip").style.paddingBottom ="1px"; document.getElementById("tooltip").style.paddingLeft ="1px"; document.getElementById("tooltip").style.paddingRight ="1px"; document.getElementById("tooltip").style.borderColor="black"; document.getElementById("tooltip").style.color="black"; document.getElementById("tooltip").style.borderWidth="1px"; document.getElementById("tooltip").style.borderStyle="solid"; document.getElementById("tooltip").style.top = screen.height; document.getElementById("tooltip").style.left = screen.height ; } note: you have to change the top and left values by adding some constant number by trial and error method eg: document.getElementById("tooltip").style.top = screen.height+500; document.getElementById("tooltip").style.left = screen.height+20 ; also check how the page looks at different resolutions. -- modified at 6:39 Tuesday 20th June, 2006