Need help for random number

Desmond Tan

Hi
can anyone teach me how to use visual basic to make a random number generate between 1 to 10

than

example:
if the generate number is 1 it will display "abc" on the textbox
if the generate number is 2 it will display "cba" on the textbox

Pls help me thank!

4 Replies

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Solitaire20y ago
Incorrect.

intRandom = rnd.Next(0, 10)

will generate a random number between 0 and 9. The second argument after Next must be 1 more than the upper limit of the range.

For the above example, you should use:

intRandom = rnd.Next(1, 11)
Scott Lysle20y ago
Starting with a single form with one button (Button1) and one textbox (TextBox1), the click event handler will generate a random number between one and ten and will use that number is a select case statement to set the text to some value associated with a number between one and ten:

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click

     Dim rnd As New Random()
     Dim intRandom As Integer
     intRandom = rnd.Next(0, 10)

     Select Case intRandom
          Case 1
               TextBox1.Text = "ABC"
          Case 2
               TextBox1.Text = "DEF"
          Case 3
               TextBox1.Text = "GHI"
          Case 4
               TextBox1.Text = "JKL"
          Case 5
               TextBox1.Text = "MNO"
          Case 6
               TextBox1.Text = "PQR"
          Case 7
               TextBox1.Text = "STU"
          Case 8
               TextBox1.Text = "VXY"
          Case 9
               TextBox1.Text = "ZAB"
          Case 10
               TextBox1.Text = "ZAC"
          Case Else
               TextBox1.Text = "AAA"
     End Select
End Sub
Desmond Tan20y ago
eer i still dun get it le... can show me the full source code please... i really need help for it...

if can send me a sample pls... [email protected]...
Scott Lysle20y ago

Dim rnd As New Random()

Dim intRandom As Integer

intRandom = rnd.Next(0, 10)

MessageBox.Show("Random Number between 0 and 10: " & intRandom.ToString())