I am trying to get my update button to function from my datagrid. But I
keep getting a index out of range error. I see this post alot, and now I'm
proabably more confused about tha ever. When I hit my edit button, it loads
the right fields into my textboxes, there I can change the info, then hit
save. But then the error pop's up. Now I can add a new record with out any
problem. I do not have a textbox for my ID column, which is my key.
To my understanding, it's not reading the "key" column properly.
I guess i'm not sure what I'm looking for. what would be the line of code
that would cause this problem, or what areas may I look at to make sure I
have it set up right. For all I know, I may not even have the right line of
code in there.
I tried the EditItemIndex = e.item thing, but I'm not sure where to put it.
I'm sorry I don't have my code with me right now, but I can post later
tonite if needed. I know it makes it alot easier to figure out.
As always, thanks!
RudyRudy,
I think it's probably best to post your code. The error typically means
you're trying to update a row (item) but the index or row positions that
you've set is greater than the total number of rows.
Jon
"Rudy" wrote:
> Hello all!
> I am trying to get my update button to function from my datagrid. But I
> keep getting a index out of range error. I see this post alot, and now I'm
> proabably more confused about tha ever. When I hit my edit button, it loads
> the right fields into my textboxes, there I can change the info, then hit
> save. But then the error pop's up. Now I can add a new record with out any
> problem. I do not have a textbox for my ID column, which is my key.
> To my understanding, it's not reading the "key" column properly.
> I guess i'm not sure what I'm looking for. what would be the line of code
> that would cause this problem, or what areas may I look at to make sure I
> have it set up right. For all I know, I may not even have the right line of
> code in there.
> I tried the EditItemIndex = e.item thing, but I'm not sure where to put it.
> I'm sorry I don't have my code with me right now, but I can post later
> tonite if needed. I know it makes it alot easier to figure out.
> As always, thanks!
> Rudy
Hi Jon!
Here is the code that I have.
Private Sub SaveItem()
Dim strSQL As String
If btnSave.CommandArgument = "Add" Then
strSQL = _
"INSERT INTO IMSProducts " & _
" (ProductName, SupplierPart, UnitCost, UnitsInStock, " & _
" Discontinued, UnitSRP, Used) " & _
"VALUES " & _
" (@.ProductName, @.SupplierPart, @.UnitCost, @.UnitsInStock,
" & _
" @.Discontinued, @.UnitSRP, @.Used)"
Else ' The user is updating an existing item.
strSQL = _
"UPDATE IMSProducts " & _
"SET ProductName = @.ProductName, " & _
" SupplierPart = @.SupplierPart, " & _
" UnitSRP = @.UnitSRP, " & _
" UnitCost = @.UnitCost, " & _
" UnitsInStock = @.UnitsInStock, " & _
" Discontinued = @.Discontinued, " & _
" Used = @.Used " & _
"WHERE ProductID = @.ProductID"
End If
Dim Sqlconnection1 As New SqlConnection(SQL_CONNECTION_STRING)
Dim scmd As New SqlCommand(strSQL, Sqlconnection1)
' Add all the required SQL parameters.
With scmd.Parameters
' The ProductID parameter is only needed for updating.
If btnSave.CommandArgument <> "Add" Then
.Add(New SqlParameter("@.ProductID", _
SqlDbType.Int)).Value = _
CInt(grdProducts.DataKeys(grdProducts.SelectedInde x).ToString)
' grdProducts.EditItemIndex = -1
End If
.Add(New SqlParameter("@.ProductName", _
SqlDbType.NVarChar, 40)).Value = txtProductName.Text
.Add(New SqlParameter("@.SupplierPart", _
SqlDbType.NVarChar, 50)).Value = txtSupplier.Text
.Add(New SqlParameter("@.UnitCost", _
SqlDbType.Money)).Value = CDbl(txtCost.Text)
.Add(New SqlParameter("@.UnitSRP", _
SqlDbType.Money)).Value = CDbl(txtSRP.Text)
.Add(New SqlParameter("@.UnitsInStock", _
SqlDbType.Int)).Value = CInt(txtInStock.Text)
.Add(New SqlParameter("@.Discontinued", _
SqlDbType.Bit)).Value = chkDiscontinued.Checked
'.Add(New SqlParameter("@.New", _
' SqlDbType.Bit)).Value = chkNew.Checked
.Add(New SqlParameter("@.Used", _
SqlDbType.Bit)).Value = chkUsed.Checked
End With
Try
Sqlconnection1.Open()
scmd.ExecuteNonQuery()
Cache.Remove("dvProducts")
BindProductsGrid()
strMsg = "Your a star! Product successfully saved to the
database."
pnlForm.Visible = True
Catch exp As Exception
strErrorMsg = "Database error! Product not saved to the " & _
"database. Error message: " & exp.Message
Finally
Sqlconnection1.Close()
End Try
End Sub
Here is the code for the save btn.
' This routine handles the "Save Changes" button Click event.
Private Sub btnSave_Click(ByVal sender As Object, ByVal e As
System.EventArgs) Handles btnSave.Click
If IsValid Then
SaveItem()
txtProductName.Text = ""
txtSupplier.Text = ""
txtCost.Text = ""
txtSRP.Text = ""
txtInStock.Text = ""
chkDiscontinued.Checked = False
chkUsed.Checked = False
End If
End Sub
I hope this makes a little more sense.
Thank you for your time!!
Rudy
"Jon" wrote:
> Rudy,
> I think it's probably best to post your code. The error typically means
> you're trying to update a row (item) but the index or row positions that
> you've set is greater than the total number of rows.
> Jon
> "Rudy" wrote:
> > Hello all!
> > I am trying to get my update button to function from my datagrid. But I
> > keep getting a index out of range error. I see this post alot, and now I'm
> > proabably more confused about tha ever. When I hit my edit button, it loads
> > the right fields into my textboxes, there I can change the info, then hit
> > save. But then the error pop's up. Now I can add a new record with out any
> > problem. I do not have a textbox for my ID column, which is my key.
> > To my understanding, it's not reading the "key" column properly.
> > I guess i'm not sure what I'm looking for. what would be the line of code
> > that would cause this problem, or what areas may I look at to make sure I
> > have it set up right. For all I know, I may not even have the right line of
> > code in there.
> > I tried the EditItemIndex = e.item thing, but I'm not sure where to put it.
> > I'm sorry I don't have my code with me right now, but I can post later
> > tonite if needed. I know it makes it alot easier to figure out.
> > As always, thanks!
> > Rudy
