AliChemicali
Legacy Member
Hello!
I'm trying to find a code to delete multiple columns because when you delete a column the next column takes his place so for example:
If you delete column 6 then column 7 becomes column six.
I tried the following code with an for loop & a while loop:
Dim field As DataControlField = gv.Columns(i)
gv.Columns.Remove(field)
&
gv.columns.removeat(i)
But same case that the column 7 becomes column 6
The variable removeColumns is an array with the column number that need to be deleted.
If wanted I can also provide the Header Names
Thanks a lot for anyone who can Help me!
I'm trying to find a code to delete multiple columns because when you delete a column the next column takes his place so for example:
If you delete column 6 then column 7 becomes column six.
I tried the following code with an for loop & a while loop:
Dim field As DataControlField = gv.Columns(i)
gv.Columns.Remove(field)
&
gv.columns.removeat(i)
But same case that the column 7 becomes column 6
The variable removeColumns is an array with the column number that need to be deleted.
If wanted I can also provide the Header Names
Thanks a lot for anyone who can Help me!
Code:
Imports System.Data
Imports System.Configuration
Imports System.IO
Imports System.Web
Imports System.Web.Security
Imports System.Web.UI
Imports System.Web.UI.WebControls
Imports System.Web.UI.WebControls.WebParts
Imports System.Web.UI.HtmlControls
Public Class GridViewExportUtil
Public Shared Sub Export(ByVal fileName As String, ByVal gv As GridView, ByVal removeColumns As Integer())
HttpContext.Current.Response.Clear()
HttpContext.Current.Response.AddHeader("content-disposition", String.Format("attachment; filename={0}", fileName))
HttpContext.Current.Response.ContentType = "application/ms-excel"
'Gridview column delete code should come here
Using sw As New StringWriter()
Using htw As New HtmlTextWriter(sw)
' Create a form to contain the grid
Dim table As New Table()
' add the header row to the table
If gv.HeaderRow IsNot Nothing Then
GridViewExportUtil.PrepareControlForExport(gv.HeaderRow)
table.Rows.Add(gv.HeaderRow)
End If
' add each of the data rows to the table
For Each row As GridViewRow In gv.Rows
GridViewExportUtil.PrepareControlForExport(row)
table.Rows.Add(row)
Next
' add the footer row to the table
If gv.FooterRow IsNot Nothing Then
GridViewExportUtil.PrepareControlForExport(gv.FooterRow)
table.Rows.Add(gv.FooterRow)
End If
' render the table into the htmlwriter
table.RenderControl(htw)
' render the htmlwriter into the response
HttpContext.Current.Response.Write(sw.ToString())
HttpContext.Current.Response.[End]()
End Using
End Using
End Sub