Hi.

Now I am tryting to open ajax popup by using onmouseover of Label control.

I added onmouseover and onmouseout attribute to label control in gridview itemtemplate.

 If e.Row.RowType = DataControlRowType.DataRow Then
            Dim lblName As Label = CType(e.Row.FindControl("lblFullName"), Label)
            lblName.Attributes.Add("onmouseover", "ShowTooltip()")
            lblName.Attributes.Add("onmouseout", "HideTooltip()")
End If

Private Sub ShowTooltip()

modalpopup.show()

End Sub 

private Sub HideTooltip()

modalpopup.hide()

End Sub

But I couldn't get any response.

Is there any way to show ajax modalpopup by using onmouseover and onmouseout attribute of label control ?

Hi

F.Y.I

I know you want add the attributes to every labels in your itemtemplate

but i think there is something wrong in the way you find the label

CType(e.Row.FindControl("lblFullName"), Label) //try to use client name instead of "lblFullName" please

Hope this helps

Hi. BlodFox

How to use client name of lable control ?

Could you show me the simple code ?

 thanks

Hi

do a ViewSource to your web page

 you can see the controls in the itemtemplete

their client names and id is defferent from their server name

for example:

a CheckBox1 in GridView1 's templete column will be

<input id="GridView1_ctl01_CheckBox1" type="checkbox" name="GridView1$ctl01&CheckBox1"/>

<input id="GridView1_ctl02_CheckBox1" type="checkbox" name="GridView1$ctl02&CheckBox1"/>

<input id="GridView1_ctl03_CheckBox1" type="checkbox" name="GridView1$ctl03&CheckBox1"/>

…………

so, you can find the label by the client name or id 

Hope this helps