[zany315]

Hello I am a new programmer in visual basics. Right now I am stomped with this recent assignment.

Write a program to play a lottery game. The player will select 3 numbers between 1 and 9; these
numbers do not have to be unique. For example, the player may select 3, 2, 9 or 6, 2, 2. The winning
numbers are randomly generated by the computer. If all three of the numbers selected by the player are
the same as the winning numbers generated, then the player wins, otherwise the player loses. The
player’s numbers and the winning numbers must be an exact match (the order the numbers appear must
match).

And so far I have:

Dim num1 As Integer
        Dim num2 As Integer
        Dim num3 As Integer

        num1 = txtnum1.Text
        num2 = txtnum2.Text
        num3 = txtnum3.Text
        num1 = CInt(txtnum2.Text)
        num2 = CInt(txtnum2.Text)
        num3 = CInt(txtnum3.Text)

        Dim randomNum As New Random 'creates a random number generator
        Dim winningNum1 As Integer
        winningNum1 = randomNum.Next(1, 9)
        Dim winningNum2 As Integer
        winningNum2 = randomNum.Next(1, 9)
        Dim WinningNum3 As Integer
        WinningNum3 = randomNum.Next(1, 9)

I am having difficulties converting the numbers to string. Also making a textbox visible after the button has been clicked. Please help

This post has been edited by modi123_1: 05 December 2010 - 11:45 AM
Reason for edit:: please use the code tags..

[CruNcK]

I also have another remark, if you generate a number with a random, and you typ (1,9), it'll generate numbers from 1 to 8, because it is exclusive upperbount

[zany315]

Thanks.

winningNum1.visible = True It says this is not integer. How would I convert this to an integer. I am really having trouble converting may someone explain that to me?

[zany315]

The latest code:

  'Written by Gerardo Morillo
        Dim num1 As Integer
        Dim num2 As Integer
        Dim num3 As Integer

        num1 = txtnum1.Text
        num2 = txtnum2.Text
        num3 = txtnum3.Text
        num1 = CInt(txtnum2.Text)
        num2 = CInt(txtnum2.Text)
        num3 = CInt(txtnum3.Text)
        num1.ToString()
        num2.ToString()

        Dim randomNum As New Random 'creates a random number generator
        Dim winningNum1 As Integer
        winningNum1 = randomNum.Next(1, 10)
        Dim winningNum2 As Integer
        winningNum2 = randomNum.Next(1, 10)
        Dim WinningNum3 As Integer
        WinningNum3 = randomNum.Next(1, 10)

        winningNum1.visible = True
        winningNum1.ToString(winningNum1)
        

        If winningNum1 = num1 Then
            num1 = True

And from here I am totally lost!

MOD EDIT: When posting code...USE CODE TAGS!!!

This post has been edited by JackOfAllTrades: 06 December 2010 - 09:01 AM

[CharlieMay]

This is unnecessary

num1 = txtnum1.Text
num2 = txtnum2.Text
num3 = txtnum3.Text

It may not throw an error with Option Strict Off. It will definitely throw an error with Option Strict On as you are trying to make an integer = a string and you can only do this if the string can be converted to an integer.

This is ok

num1 = CInt(txtnum2.Text)
num2 = CInt(txtnum2.Text)
num3 = CInt(txtnum3.Text)

But you're making a huge assumption that the user is going to follow instruction and enter a number in the textbox that can be cast to an integer. Either use a numericupdown control to force the user to only use a number or use .TryParse to test the textboxes.

Not sure why you're doing this as it is doing nothing at all for you.
num1.ToString()
num2.ToString()


Try removing

num1 = txtnum1.Text
num2 = txtnum2.Text
num3 = txtnum3.Text

and

num1.ToString()
num2.ToString()

Now this:

winningNum1.visible = True
winningNum1.ToString(winningNum1)

Will not work because an integer has no visible property. If you want to display the result of winningNum1 you could add a label and set its .Text property to winningNum1.Tostring()
This will show what the number generated from the random is.
The second line even if it was set up correctly would not do anything much like the num1.tostring() line above.
Remove both of thes lines.

The last thing you have is num1 = true. num1 is an integer and true is a boolean result. You need to use an if statement to check for true and then display when it is:

Example:

If winningNum1 = num1 then
  MessageBox.Show("You matched number 1 to a random number")
End If

Keep in mind, you will only see the messagebox IF the number typed into textbox1 matches the random number generated for winningNum1. So you would have to do this for each of the winningnum(s) and num(s) variables. Either by adding additional If...Then Blocks or using ElseIf for the other comparisons.

see if that gets you closer to your goal and then we can help you use methods appropriate to the class assignment.

This post has been edited by CharlieMay: 06 December 2010 - 06:58 AM

[zany315]

I am still confused about the conversion process.

Dim num1 As Integer = CStr(txtnum1.Text)
Dim num2 As Integer = CStr(txtnum2.Text)
Dim num3 As Integer = CStr(txtnum3.Text)
Dim messagebox As String
Dim winningNum1 As Integer
Dim winningnum2 As Integer
Dim winningnum3 As Integer

This part here is my biggest problem. I am not sure what to do here. How do you make it so that the conversion from int to string is acceptable?

MOD EDIT: When posting code...USE CODE TAGS!!!

This post has been edited by JackOfAllTrades: 06 December 2010 - 09:02 AM

[CruNcK]

do it like this:

dim num1, num2, num3, winningNum1, winningNum2, winningNum3 as integer

num1 = integer.parse(txtnum1.text)
num2 = integer.parse(txtnum2.text)
num3 = integer.parse(txtnum3.text)

[begbodyguard]

'Declaration

'User numbers
Dim intnum1 As Integer = CInt(txtnum1.Text)
Dim intnum2 As Integer = CInt(txtnum2.Text)
Dim intnum3 As Integer = CInt(txtnum3.Text)

'Winning numbers (pc)
Dim intwinningNum1 As Integer
Dim intwinningnum2 As Integer
Dim intwinningnum3 As Integer

'Booleans to store if number1 = winningnumber and number2 .....
Dim bliscorrect1 as boolean
Dim bliscorrect2 as boolean
Dim bliscorrect3 as boolean
'Make the pc choose a random number
Dim randomx as new random

Changed it a little bit start too with
Hope this will help you

[zany315]

Thanks all for the help. It really did help!