每次我生成一個字符串，然後關閉該程序，並再次生成它相同的

Question

嗯，我做了這個隨機字符串生成，但我最近注意到一個致命的缺陷。當我生成一個隨機字符串時，關閉該程序並生成另一個隨機字符串，它與第一代相同。下面是代碼：

Public Function RandomString(ByVal length As Integer) As String 
    Dim strb As New System.Text.StringBuilder 
    Dim chars() As String = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"} 

    Dim UpperBound As Integer = UBound(chars) 

    For x As Integer = 1 To length 
     strb.Append(chars(Int(Rnd() * UpperBound))) 
    Next 

    Return strb.ToString 

End Function 

Private Sub PictureBox1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles PictureBox1.Click 
    Try 
     System.Diagnostics.Process.Start("Link Removed!") 
    Catch 
    End Try 
End Sub 

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

    'Check for valid numeric entry 
    If Integer.TryParse(TextBox2.Text, repeatCnt) Then 
     For repeatIdx As Integer = 1 To repeatCnt 
      Dim rndstring As String 
      'Generate random string... 
      rndstring = RandomString(24) 
      '...and append to text box with a line break 
      RichTextBox1.Text &= rndstring & vbCrLf 
     Next 
    Else 
     MessageBox.Show("Please enter a valid integer number in the text box") 
    End If 
End Sub 

Answer 1

你的問題是，你正在運行賽第一輪（）而不是使用Random類。

爲了保持Rnd()不會每次給出相同的「隨機」數字序列，應該調用Random.Next()（爲了使用機器時鐘作爲初始值/種子）。

所以你的情況：

Dim random As New System.Random() 
Public Function RandomString(ByVal length As Integer) As String 
    Dim strb As New System.Text.StringBuilder 
    Dim chars() As String = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"} 

    Dim UpperBound As Integer = UBound(chars) 

    For x As Integer = 1 To length 
     strb.Append(chars(Int(random.Next(UpperBound))) 
    Next 

    Return strb.ToString 

End Function  

Answer 2

嘗試生成一個隨機字符串這樣的：

Public Shared Function GeneratePass() As String 

    Dim unique As Guid = Guid.NewGuid 
    Return unique.ToString.Substring(1, 6) 

End Function