我如何验证这一点？

Question

I have made this validation and I know it is very inefficient but the validation isn't completely working. What I mean is, when I was testing it I realised if I fail the first condition, then fix it, then fail the second condition, then when I fix the input to work with the second condition, I can make the input not correct for the first condition but it still passes through. I might have explained that very wrong but here is my code:

Function NewPassword(ByRef New_password As String)

    Dim Valid1, Valid2 As Boolean
    Dim First_char As String
    Dim last_char As String
    Dim First_char_asc As Integer
    Dim last_char_asc As Integer


    Valid1 = False
    Valid2 = False

    Do Until Valid1 = True And Valid2 = True
        Do Until Valid1 = True

            Valid1 = False

            New_password = InputBox("New password needs a capital letter at the start, Re-enter a new password.")

            First_char = Mid$(New_password, 1, 1)
            First_char_asc = Asc(First_char)


            If First_char_asc >= 65 Or First_char_asc <= 90 Then

                Valid1 = True

            End If

        Loop

        Do Until Valid2 = True

            Valid2 = False

            New_password = InputBox("New password needs a symbol at the end, Re-enter a new password.")

            last_char = Mid$(New_password, Len(New_password), 1)
            last_char_asc = Asc(last_char)

            If last_char_asc >= 35 Or last_char_asc <= 37 Then

                Valid2 = True

            End If

        Loop

        If Valid1 = False Or Valid2 = False Then
            Valid1 = False And Valid2 = False
        End If
    Loop

    Return New_password

End Function

If you recognise this problem or you can help me out in my situation please respond.

Answer 1

you can use single validation blog for both the validation. Since I am using mobile to answer please ignore the mess with the format.

Function NewPassword(ByRef New_password As String)
    Dim Valid1, Valid2 As Boolean
    Dim First_char As String
    Dim last_char As String
    Dim First_char_asc As Integer
    Dim last_char_asc As Integer
    Valid1 = False
    Valid2 = False
    Do Until Valid1 = True
        Valid1 = False
        New_password = InputBox("New password needs a capital letter at the start and a symbol at the end. Re-enter a new password.")
        First_char = Mid$(New_password, 1, 1)
        First_char_asc = Asc(First_char)
        last_char = Mid$(New_password, Len(New_password), 1)
        last_char_asc = Asc(last_char)
        If (First_char_asc >= 65 and First_char_asc <= 90) And (last_char_asc >= 35 and last_char_asc <= 37) Then
            Valid1 = True
        End If
    Loop
    Return New_password
End Function

Answer 2

Perhaps you could do something along these lines...

Function ValidateCapital(New_password, valid1)
    Dim First_char = Mid$(New_password, 1, 1)
    Dim First_char_asc = Asc(First_char)
    If First_char_asc >= 65 And First_char_asc <= 90 Then
        valid1 = True
    Else
        valid1 = False
    End If
    Return valid1
End Function

Function ValidateSymbol(New_password, valid2)
    Dim last_char = New_password(New_password.Length - 1)
    Dim last_char_asc = Asc(last_char)
    If last_char_asc >= 35 And last_char_asc <= 37 Then
        valid2 = True
    Else
        valid2 = False
    End If
    Return valid2
End Function

Function NewPassword(ByRef New_password As String)
    Dim Valid1, Valid2 As Boolean
    Valid1 = ValidateCapital(New_password, Valid1)
    Valid2 = ValidateSymbol(New_password, Valid2)
    While Valid1 <> True Or Valid2 <> True
        If Valid1 = False Then
            New_password = InputBox("New password needs a capital letter at the start, Re-enter a new password.")
            Valid1 = ValidateCapital(New_password, Valid1)
        End If
        If Valid1 = True Then
            Valid2 = ValidateSymbol(New_password, Valid2)
            If Valid2 = False Then
                New_password = InputBox("New password needs a symbol at the end, Re-enter a new password.")
                Valid2 = ValidateSymbol(New_password, Valid2)
            End If
        End If
        Valid1 = ValidateCapital(New_password, Valid1)
        Valid2 = ValidateSymbol(New_password, Valid2)
    End While
    Return New_password
End Function

Or a simpler version could be made from something like...

Sub Main()
    Dim valid1 = False
    Dim valid2 = False
    While valid1 = False Or valid2 = False
        Dim Password = InputBox(“Enter your password”)
        Dim First_char = Asc(Mid$(Password, 1, 1))
        If valid1 = True And (First_char >= 65 Or First_char <= 90) Then
            valid1 = True
        End If
        Dim Last_char = Asc(Mid$(Password, Len(Password), 1))
        If valid2 = True And (Last_char >= 35 Or Last_char <= 37) Then
            valid2 = True
        End If
    End While
End Sub

Answer 3

What are you returning from your password? You must declare a type for your function.

Mid went out with VB6 although it is still included for backward compatibility. Use SubString in new code.

Those Do loops are begging for endless loops if the User doesn't comply or gives up.

Your password rules would be helpful to a hacker. Are you salting and encrypting before storage?

Instead of an InputBox use a TextBox . Then you can use the Validating event to test the password.

Fortunately a String is an array of Char . The Char structure has many methods to determine what kind of character it is.

Private Sub TextBox1_Validating(sender As Object, e As System.ComponentModel.CancelEventArgs) Handles TextBox1.Validating
    Dim ValidString = ValidatePassword(TextBox1.Text)

    If Not ValidString = "Valid" Then
        MessageBox.Show(ValidString)
        e.Cancel = True
    End If
End Sub

Private Function ValidatePassword(New_password As String) As String
    If Char.IsLower(New_password(0)) Then
        Return "New password needs a capital letter at the start, Re-enter a new password."
    End If
    If Not Char.IsSymbol(New_password(New_password.Length - 1)) Then
        Return "New password needs a symbol at the end, Re-enter a new password."
    End If
    Return "Valid"
End Function

Answer 4

They teaching you vb6 stuff? Show your teacher/professor the following:

Public Function ValidatePassword(Pw as string, byref ErrorMessage As String) As Boolean
ErrorMessage = ""
Dim FirstValidation As Boolean, SecondValidation As Boolean
Dim firstChar as char = Pw.SubString(0,1)
Dim lastChar as char = Pw.SubString(pw.length,1)

FirstValidation = asc(firstchar) >= 65 And asc(firstchar) <= 90
If FirstValidation = False Then ErrorMessage = "The first character is not Uppercase" & vbnewline
SecondValidation = asc(lastchar) >= 35 And last_char_asc <= 37
If SecondValidation = False Then ErrorMessage &= "The last character is not a symbol" & vbnewline
Return FirstValidation And SecondValidation
End Function

I don't know why you have to be in an endless loop like that. You pass this function in your button.

ButtonClicK:

Dim errorMessage as string = String.Empty
If ValidatePassword(textbox1.text, ErrorMessage) = False then msgbox.Show(errorMessage)