Question

我们上次有一项活动。它将我同学的python代码转换为vb ...这是我的最终代码，它正在运行。

Private Sub txtInput_KeyPress(KeyAscii As Integer)
    If KeyAscii = 13 Then
    curyear = Int(2013)
    a = Int((curyear - txtInput.Text) Mod 12)



        txtInput.Text = " "
        If (a = 9) Or (a = -3) Then
            txtOutput.Text = "Your zodiac sign is Snake"
        ElseIf (a = 8) Or (a = -4) Then
            txtOutput.Text = "Your zodiac sign is Dragon"
        ElseIf (a = 7) Or (a = -5) Then
            txtOutput.Text = "Your zodiac sign is Rabbit"
        ElseIf (a = 6) Or (a = -6) Then
            txtOutput.Text = "Your zodiac sign is Tiger"
        ElseIf (a = 5) Or (a = -7) Then
            txtOutput.Text = "Your zodiac sign is Ox"
        ElseIf (a = 4) Or (a = -8) Then
            txtOutput.Text = "Your zodiac sign is Rat"
        ElseIf (a = 3) Or (a = -9) Then
            txtOutput.Text = "Your zodiac sign is Pig"
        ElseIf (a = 2) Or (a = -10) Then
            txtOutput.Text = "Your zodiac sign is Dog"
        ElseIf (a = 1) Or (a = -11) Then
            txtOutput.Text = "Your zodiac sign is Rooster"
        ElseIf (a = 0) Or (a = -2) Then
            txtOutput.Text = "Your zodiac sign is Monkey"
        ElseIf (a = 11) Or (a = -1) Then
            txtOutput.Text = "Your zodiac sign is Sheep"
        ElseIf (a = 12) Or (a = 0) Then
            txtOutput.Text = "Your zodiac sign is Horse"
        End If
    End If

    End Sub

我们的教授告诉我其他使用ascii 13 /进入......我无法理解他。您认为我的代码有什么问题？它正在运行，但他说我的代码错了。

Answer 1

首先，稍微清理你的代码。

不是使用所有那些Else If语句，而是采用更好的方法使用Select Case。
您的第3或第4行没有Int（）。它已经是一个整数。
你也不必重复“你的生肖是”的字符串。只需使用一次。
正如Deanna所说，您需要从文本框中删除ASCII 13 输入

所以，新代码：

Private Sub txtInput_KeyPress(KeyAscii As Integer)
   Dim ZodiacAnimal As String
   If KeyAscii = 13 Then
      curyear = 2013
      a = (curyear - txtInput.Text) Mod 12
      Select Case a
         Case 9, -3
            ZodiacAnimal = "Snake"
         Case 8, -4
            ZodiacAnimal = "Dragon"
         Case 7, -5
            ZodiacAnimal = "Rabbit"
         Case 6, -6
            ZodiacAnimal = "Tiger"
         Case 5, -7
            ZodiacAnimal = "Ox"
         Case 4, -8
            ZodiacAnimal = "Rat"
         Case 3, -9
            ZodiacAnimal = "Pig"
         Case 2, -10
            ZodiacAnimal = "Dog"
         Case 1, -11
            ZodiacAnimal = "Rooster"
         Case 0, -2
            ZodiacAnimal = "Monkey"
         Case 11, -1
            ZodiacAnimal = "Sheep"
         Case 12, 0
            ZodiacAnimal = "Horse"
      End Select
      txtInput.Text = "Your zodiac sign is " & ZodiacAnimal
      KeyAscii to 0
   End If
End Sub

现在，我看到了其他一些问题。

你有一年硬编码。十二生肖大约是你出生的那一年，与今年没有任何关系。
你有两次列出。

现在，这对你来说可能太先进了，但这就是我对它的编码方式：

Private Sub txtInput_KeyPress(KeyAscii As Integer)
   Dim ZodiacAnimal() As String
   If KeyAscii = 13 Then
      ZodiacAnimal = Split("Monkey,Rooster,Dog,Pig,Rat,Ox,Tiger,Rabbit,Dragon,Snake,Horse,Goat", ",")
      txtInput.Text = "Your zodiac sign is " & ZodiacAnimal(Y Mod 12) 
      KeyAscii to 0
   End If
End Sub

这使得ZodiacAnimal成为动物的一个阵列。然后使用Mod函数，我得到了数组的正确索引。

Answer 2

另外，使用

的方式更为合适

vbKeyReturn

而不是数字13，因为并非所有键盘都是ENTER。

所以正确的方法是：

If KeyAscii = vbKeyReturn Then

（请评价我的回答，谢谢！）

我的计划有什么问题？关于Ascii 13 [Enter]

2 个答案: