[RESOLVED] [2008] - Datagridview to multidimensional array

Hello

I need some advice to accomplish this, i have a datagridview with this data:
Column1 Column2
item1 3
item2 1
item1 2
item1 5
item2 3
item3 1
item1 3

I need an array with distinct values of column1 and the sum of the values in the second dimension of the array:
ar(item1,13), ar(item2,4),ar(item3,1).

I don't know how many items i will have, so i don't know the correct dimension of the array...

I have a lot of code to do the job, but i think there is a better way...

Thanks

Is there a datatable bound to the source of the datagridview?

If so, you can run sql like commands against it.

i.e

Code:

---

      Dim dt As New DataTable
        Dim col1Sum As Integer = dt.Compute("SUM(DISTINCT Column1)", "")

---

No there isn't... :(

Well, I don't think you really do want a 2D array, or at least you shouldn't. There are some, but very few, situations where 2D arrays are appropriate. They should only be used when every element is a peer. If one column represents something completely different to another column, as is the case here, the a 2D array is completely inappropriate.

One appropriate solution would be to use a Dictionary:

vb.net Code:

Dim itemSums As New Dictionary(Of String, Integer)
Dim item As String
 
For Each row As DataGridViewRow In Me.DataGridView1.Rows
    item = CStr(row.Cells(0).Value)
 
    If itemSums.ContainsKey(item) Then
        'Add the new value to the current subtotal.
        itemSums(item) += CInt(row.Cells(1).Value)
    Else
        'Add the item and the first value.
        itemSums.Add(item, CInt(row.Cells(1).Value))
    End If
Next

Another option would be to define a type that represents an item and its corresponding value and to put instances of that type in a simple collection. In both cases you're covered for the unknown number of items because collections can grow dynamically.

If you really must use a 2D array, like if this is a poor learning exercise, then you can create the array from the collection, because at that point you know how many elements you need.

THanks again :)

I was looking for a code similar to that a long time...

Solved