嗨,我有这个代码

Dim RandomNumber As New Random()
Dim msg As String = "My name is Nick from Neptune"
For each space as char in msg
msg = msg.replace(" ", RandomNumber.Next(0, 33))
Next

但这是行不通的。
有人能帮我吗?
最后,字符串应如下所示:
My12name455is22Nick88from66Neptune

在此先感谢您,我的英语很抱歉。

最佳答案

我改变了它的工作,我很抱歉:

Option Infer On

Module Module1

Sub Main()
    Dim RandomNumber As New Random()
    Dim msg As String = "My name is Nick from Neptune"
    Dim newmessage As String = ""
    Dim d = msg.Split(" ").ToList()

    For Each t In d
        newmessage = newmessage & Convert.ToString(t) & RandomNumber.[Next](0, 33).ToString()
    Next

    Console.WriteLine(newmessage)
End Sub

终端模块

10-07 19:46
查看更多