Data Presentation Controls

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

someguy198650 — Thu, 12 Jun 2008 02:46:32 GMT

gal12:
thanks.. i managed to solve my problem.. i missed out one part of the code... opps....

hi gal12

can you show me that code??

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

someguy198650 — Thu, 12 Jun 2008 02:24:42 GMT

hi joteke,

i have the same freaking problem, i cant use the e.commandargument because im assigning some other values to it, like the ID in my table, somehow i want to find the selected index in my gridview.

here is my post http://forums.asp.net/t/1273810.aspx, can you help me?

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

scorpion1 — Wed, 11 Jun 2008 11:47:03 GMT

hi guys need some help i added some insert cods but there something WRONG I’m getting duplicate rows the table

Protected Sub GridView1_RowCommand(ByVal sender As Object, ByVal e As System.Web.UI.WebControls.GridViewCommandEventArgs) Handles GridView1.RowCommand

If e.CommandName = "myRowSelect" Then

Dim rowindex As Integer = CInt(e.CommandArgument)

Dim row As GridViewRow = GridView1.Rows(rowindex)

Dim ItemnameLabel As Label = CType(GridView1.Rows(rowindex).FindControl("ItemnameLabel"), Label)

Dim ItempriceLabel As Label = CType(GridView1.Rows(rowindex).FindControl("ItempriceLabel"), Label)

Dim itemcountextbox As TextBox = CType(GridView1.Rows(rowindex).FindControl("itemcountextbox"), TextBox)

itempricelabel2.Text = ItempriceLabel.Text

itemnamelabel2.Text = ItemnameLabel.Text

itemcountlabe2.text = itemcountextbox.Text

End If

Using myConnection As New SqlConnection(ConfigurationManager.ConnectionStrings("ConnectionString").ConnectionString)

Const SQL As String = "INSERT INTO [cart] ([itemprice], [itemname],[itemcount]) VALUES (@itemprice, @itemname , @itemcount)"

Dim myCommand As New SqlCommand(SQL, myConnection)

myCommand.Parameters.AddWithValue("@itemname", itemnamelabel2.Text.Trim())

myCommand.Parameters.AddWithValue("@itemprice", itempricelabel2.Text.Trim())myCommand.Parameters.AddWithValue("@itemcount", itemcountlabe2.Text.Trim())

myConnection.Open()

myCommand.ExecuteNonQuery()

myConnection.Close()

End Using

End Sub

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

wantedwahab — Fri, 30 May 2008 11:12:05 GMT

HotFrost:

i need to write:

Dim row_to_insert as GridViewRow = GridView1.FooterRow..:):):) - there is no index for footer - just footer row..

But i still can't get those textboxes in FooterRow to give me their values to make a record insertion!!!

this post helped me as a solution to my problem

Thank you very much

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

waskelton4 — Wed, 06 Feb 2008 17:15:26 GMT

Ok...

This is unbeleivable!

I have this code running when I click the Edit commandField column in my grid view.

1    Protected Sub gvWeights_RowCommand(ByVal sender As Object, ByVal e As System.Web.UI.WebControls.GridViewCommandEventArgs) Handles gvWeights.RowCommand
2               Select e.CommandName.ToLower
3                Case "edit"
4    
5                    Dim rowIndex As Integer = CInt(e.CommandArgument)
6                    gvWeights.SelectedIndex = rowIndex
7                    Dim row As GridViewRow = gvWeights.Rows(rowIndex)
8    
9                    Dim ddl As DropDownList = DirectCast(row.FindControl("WeighDatesDDL"), DropDownList)
10                   ddl.SelectedIndex = -1
11   
12           End Select
13   
14       End Sub

I get an "Object reference not set to an instance of an object." Error on line 10

I have had unreal troubles with the FindControl method. what gives?
anyone see what the problem is here? It is driving me nuts.
any and all help is greatly appreciated.

Will

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

anievarghese — Thu, 19 Jul 2007 10:58:17 GMT

There is a Gridview1, in which i have a Column BankCd.I have defined this column as a Template Field with Cell as TextBox and ReadOnly.

But in my Code i want to make this Field ReadWrite, depending on which row is selected, say first row.This code is not workin. PLEASE HELP. I am stuck on this for 2 days plus

My Code in VB:

Dim TextBox1 As TextBox

TextBox1 = CType(GridView1.Rows(0).FindControl("TB_BankCode"), TextBox)

TextBox1.ReadOnly = False

The column is still Readonly. Any Suggestions - Anyone??!

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

iloveny — Wed, 15 Nov 2006 15:17:50 GMT

for the line of "Dim mytextbox As TextBox = GridView1.FooterRow.FindControl("add_Name")"

Change it to:

Dim mytextbox as TextBox = Ctype(GridView1.FootherRow.FindControl("add_Name"), TextBox)

Hope that helps.

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

HotFrost — Fri, 10 Nov 2006 18:09:15 GMT

works fine now..was not related to the code thing..Thanks to everyone!!!

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

HotFrost — Fri, 10 Nov 2006 17:37:44 GMT

i think i got where i made an error:

i need to write:

Dim row_to_insert as GridViewRow = GridView1.FooterRow..:):):) - there is no index for footer - just footer row..

But i still can't get those textboxes in FooterRow to give me their values to make a record insertion!!!

Dim mytextbox As TextBox = GridView1.FooterRow.FindControl("add_Name")

Name = mytextbox.Text

What am i doing wrong? Thanks!

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

gal12 — Fri, 10 Nov 2006 17:01:39 GMT

thanks.. i managed to solve my problem.. i missed out one part of the code... opps....

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

HotFrost — Fri, 10 Nov 2006 16:21:25 GMT

Thanks God i found this thread...yet i have some problems to implement the code in it..May be someone, can help me out..

Here is the thing. I am trying to have 'InsertNewRecord' row in the footer of DataGrid:

i have <Footer Template> where i have text boxes for entering a new record. For example - one of the columns:

<asp:TemplateField HeaderText="Name">

<asp:TextBox ID="add_Name" Runat="Server" />

</FooterTemplate>

#Eval("Name")%>

</ItemTemplate>

<asp:TextBox ID="Name" Width="100%" Text='<%#Bind("Name") %>' Runat="server" />

</EditItemTemplate>

</asp:TemplateField>

as you can guess, my edit button is also located in the <footer template> but in the last column.

Now. For Insert Command i have this Sub:

Protected

Sub GridView1_RowCommand(ByVal sender As Object, ByVal e As System.Web.UI.WebControls.GridViewCommandEventArgs)

Dim row As GridViewRow = GridView1.Rows(0)

(correct me if i am wrong, but 'row' variable will point to footer row..am i wrong?

Nevertheless, i put the code i found in this thread:

Dim rowindex As Integer = CInt(e.CommandArgument)

Dim row_to_insert As GridViewRow = GridView1.Rows(rowindex)

Dim TempControl As Control = CType(row_to_insert_index.FindControl("add_Name"), Control)

Name = TempControl.ID

it gives me an error:

'system.FormatException: Input string was not in a correct format.' for Dim rowindex As Ingeger = CInt (e.CommandArgument)

What may be a problem? Thank you for any help and suggestions!!

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

iloveny — Fri, 10 Nov 2006 15:18:23 GMT

Can you paste the whole code?

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

gal12 — Wed, 08 Nov 2006 10:02:27 GMT

hello..

I tried to use this function as well.. but i got an error stating that String cannot be converted to int type on this line
" Dim rowindex As Integer = CInt(e.CommandArgument) "

I tried to convert to String and int16 ,int 32 type but to avail...

I need some help..

Thanks in advance..

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

iloveny — Sun, 11 Jun 2006 16:41:41 GMT

just want to drop a message to thank you for your help again.
I have already finished the project with modifying a little of your code. thanks JOTEKE!

Re: Frustated of FindControl. FindControl in GridView's ItemTemplate

iloveny — Sun, 11 Jun 2006 14:39:04 GMT

JOTEKE, thanks youuuu sooooo mucchhhh for your help. That's exactly what I need. Thanks...!!!