Question

Private Function Token() As String

        Dim Length As Byte
        Length = 10
        Dim Chars As Char() = New Char() {"a"c, "b"c, "c"c, "d"c, "e"c, "f"c, _
       "g"c, "h"c, "i"c, "j"c, "k"c, "l"c, _
        "m"c, "n"c, "o"c, "p"c, "q"c, "r"c, _
       "s"c, "t"c, "u"c, "v"c, "w"c, "x"c, _
        "y"c, "z"c, "A"c, "B"c, "C"c, "D"c, _
       "E"c, "F"c, "G"c, "H"c, "I"c, "J"c, _
       "K"c, "L"c, "M"c, "N"c, "O"c, "P"c, _
        "Q"c, "R"c, "S"c, "T"c, "U"c, "V"c, _
       "W"c, "X"c, "Y"c, "Z"c, "0"c, "1"c, _
       "2"c, "3"c, "4"c, "5"c, "6"c, "7"c, _
       "8"c, "9"c}

        Dim [Str] As String = String.Empty
        Dim Random As New Random()

        For a As Byte = 0 To Length - 1
            [Str] += Chars(Random.[Next](0, 61))
        Next
        Return ([Str])
    End Function

txtpassword.Text =令牌（）

以上粗体线是正确的..它也不起作用

以上代码用于为应该在文本框中显示的用户生成随机密码，如果可能的话，在标签中显示更为明智的

Answer 1

1）编写一些测试以查看您的Token方法是否有效。如果您使用的是Visual Studio，甚至可以在IDE中进行测试。在return方法的Token语句中设置断点，然后使用 Watch 窗口查看将返回的值。

2）您的txtpassword.Text = Token()声明似乎是正确的。这就提出了一个问题：txtpassword是否实际上是您正在查看的表单上的控件，并且也是可见的。是吗？（如果您动态创建了txtpassword，是否进行了myForm.Controls.Add(txtpassword)？

（你的代码还有其他一些问题。 - 例如make Length是一个常量，重命名变量，使它们不具有内置类型或关键字的名称，可能找到更好的方法初始化char数组等 - 但这是另一个问题。）

Answer 2

使用除[String]之外的任何其他变量名称，它是保留关键字，可能会产生问题，然后尝试。

Answer 3

我刚测试了你的代码，它运行正常没有任何问题，只是我在表单上放了一个文本框和一个按钮，然后将你的函数复制/粘贴到表单类中并放入

txtpassword.Text = Token()

按钮点击...

这是完整的代码：

Public Class Form1

    Private Function Token() As String

        Dim Length As Byte
        Length = 10
        Dim Chars As Char() = New Char() {"a"c, "b"c, "c"c, "d"c, "e"c, "f"c, _
       "g"c, "h"c, "i"c, "j"c, "k"c, "l"c, _
        "m"c, "n"c, "o"c, "p"c, "q"c, "r"c, _
       "s"c, "t"c, "u"c, "v"c, "w"c, "x"c, _
        "y"c, "z"c, "A"c, "B"c, "C"c, "D"c, _
       "E"c, "F"c, "G"c, "H"c, "I"c, "J"c, _
       "K"c, "L"c, "M"c, "N"c, "O"c, "P"c, _
        "Q"c, "R"c, "S"c, "T"c, "U"c, "V"c, _
       "W"c, "X"c, "Y"c, "Z"c, "0"c, "1"c, _
       "2"c, "3"c, "4"c, "5"c, "6"c, "7"c, _
       "8"c, "9"c}

        Dim [Str] As String = String.Empty
        Dim Random As New Random()

        For a As Byte = 0 To Length - 1
            [Str] += Chars(Random.[Next](0, 61))
        Next
        Return ([Str])
    End Function


    Private Sub SimpleButton1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles SimpleButton1.Click
        TextEdit1.Text = Token()
    End Sub
End Class

将函数的值传递到文本框中

3 个答案: