luthar
asked on
How to get the 3rd character in a string
Well the title says it all
Let's say i have the string: "abcdefg"
i want to get the 3rd character (ie: "c") what is the function to do so?
This shouln't be too difficult to do
i know in excel the STXT function does it
TIAFYT (thanks in advance for your time)
Luthar
Let's say i have the string: "abcdefg"
i want to get the 3rd character (ie: "c") what is the function to do so?
This shouln't be too difficult to do
i know in excel the STXT function does it
TIAFYT (thanks in advance for your time)
Luthar
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ASKER
Good comment but here's my code:
Option explicit
Dim Nbr_de_coup as integer
Private Sub Form_load()
Genere_La_Solution (Nbr_de_coup)
End sub
--------------------------
Public Function Genere_La_Solution(ByVal Nbr_de_Case)
Dim x As Integer
Dim Nbr_Genere As String
For x = 0 To Nbr_de_Case
Let Nbr_Genere = Int(Rnd() * 5)
Solution = Solution & Nbr_Genere
Next x
End Function
--------------------------
Public Sub Label1_Click(Index As Integer)
Dim i As Integer
Nbr_de_coup = Nbr_de_coup + 1
For i = 1 To Len(Solution)
Reponse = Mid(Solution, i, 1)
If Reponse <> Index Then
MsgBox "Vous avez pousser une mauvaise touche!", vbOKOnly, "Que c'est dommage!"
Exit Sub
End If
Next i
--------------------------
I get an error at the If Response <>...
Any suggestion?
Option explicit
Dim Nbr_de_coup as integer
Private Sub Form_load()
Genere_La_Solution (Nbr_de_coup)
End sub
--------------------------
Public Function Genere_La_Solution(ByVal Nbr_de_Case)
Dim x As Integer
Dim Nbr_Genere As String
For x = 0 To Nbr_de_Case
Let Nbr_Genere = Int(Rnd() * 5)
Solution = Solution & Nbr_Genere
Next x
End Function
--------------------------
Public Sub Label1_Click(Index As Integer)
Dim i As Integer
Nbr_de_coup = Nbr_de_coup + 1
For i = 1 To Len(Solution)
Reponse = Mid(Solution, i, 1)
If Reponse <> Index Then
MsgBox "Vous avez pousser une mauvaise touche!", vbOKOnly, "Que c'est dommage!"
Exit Sub
End If
Next i
--------------------------
I get an error at the If Response <>...
Any suggestion?
RESPONSE is a STRING, INDEX is an INTEGER. You can't compare an Integer to a String and get meaningful results...
M
M
Is this what you are trying to do?
'...
i = InStr(1, solution, Format(Index))
If i <> Index Then
MsgBox "Vous avez pousser une mauvaise touche!", vbOKOnly, "Que c'est dommage!"
Exit Sub
End If
'...
'...
i = InStr(1, solution, Format(Index))
If i <> Index Then
MsgBox "Vous avez pousser une mauvaise touche!", vbOKOnly, "Que c'est dommage!"
Exit Sub
End If
'...
ASKER
Sounds more like it but the only problem is that I need the result to be different than 0 or 1 cause i have 5 label in the control array...
I'm gonna put more points cause i think it's worth more :D
I'm gonna put more points cause i think it's worth more :D
To convert a integer to a string use STR$(num). To convert from a string to a number use VAL("string").
ASKER