Average number on ListView column

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Hi,

Is somebody can help me with the following question?

How can I get the AVERAGE number from the column of ListView1.Items(ListView1.Items.Count - 1).SubItems.Add((reader("High").ToString) - (reader("Low").ToString)) and shows the AVERAGE number on the label.

Dim Cmd As SqlCommand = New SqlCommand(SelectCmd, conn)
Dim reader As SqlDataReader
reader = Cmd.ExecuteReader()

'**************************
Do While (reader.Read())

ListView1.Items.Add(reader("Id").ToString)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(reade("Date").ToString)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(reade("High").ToString)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(reader("Low").ToString)
ListView1.Items(ListView1.Items.Count - 1).SubItems.Add((reader("High").ToString) - (reader("Low").ToString))

Loop
'***************************
 reader.Close()
 conn.Close()

For example:
The column outputs will be:
1
3
5
2
8
________
Avg: 3.8

That will be the number i would like to have it on the label.

Thank you for your help!

Posted 20-Apr-10 2:41am

vennesschan

Updated 20-Apr-10 6:24am

Tarakeshwar Reddy

v3

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William Winner · Answer 1 · 2010-04-20T06:19:00

This really isn't that difficult. I sure hope you're just starting out and actually in a class doing this.

You need to think through the problem logically. What do you need to do to solve it?

Well, you need to go through each item in the ListView and sum up their values.

Then, divide that value by the total number of items, right? What exactly are you unclear about?

VB

Dim sum as Long = 0

For Each lvItem as ListViewItem in ListView1.Items
  sum += CInt(item.SubItems(0).Text)
Next

Label1.Text = sum / ListView1.Items.Count

If you can't understand something as simple as that, maybe this isn't the profession for you.

Chris Meech · Answer 2 · 2010-04-20T02:50:00

Solution 1

How about

ListView1.Items(ListView1.Items.Count - 1).SubItems.Add(((reader(High)- (reader(Low)) / 2).ToString))

Or you could also post your SQL and have a new field added to the query.

Posted 20-Apr-10 2:50am

Chris Meech