Word and vowel count in VB 2017 - vb.net

Question

I need to program a simple console application that takes an input string and then calls a sub routine to figure out the amount of words and the amount of vowels in the string. I've written this and for some reason it doesn't output the text I put in the Console.Writeline code. Any help on how to do this?
Module Module1
Sub Main()
Dim Sentence As String
Console.WriteLine("Sentence Analysis")
Console.WriteLine()
Console.WriteLine("Enter a sentence then press 'Enter'")
Console.WriteLine()
Sentence = Console.ReadLine()
Sentence = Sentence.ToUpper
Call Words(Sentence)
Call Vowels(Sentence)
Console.ReadLine()
End Sub
Sub Words(ByVal Input As String)
Dim Count As Integer
Dim Index As Short
Dim Character As Char
Do Until Index = Len(Input)
Character = Input.Substring(Index)
If Character = " " Then
Count = Count + 1
End If
Loop
Console.WriteLine("Your sentence contains {0} words.", (Count))
End Sub
Sub Vowels(ByVal Input As String)
Dim Count As Integer
Dim Vowels() As String = {"A", "E", "I", "O", "U"}
Dim Index As Short
Dim Character As Char
Do Until Index = Len(Input)
Character = Input.Substring(Index)
If Character = Vowels(Index) Then
Count = +1
End If
Loop
Console.WriteLine("Your sentence contains {0} words.", (Count))
End Sub
End Module

Answer 1

In Words you have the following code:
Do Until Index = Len(Input)
Index is never incremented and so loops infinitely.
Same issue in Vowels as well

Answer 2

Is the following any help to you?
Rather than looking at whether a sentence contains vowels, it looks at whether a list of vowels contains each character in the sentence.
For the word count, it similarly asks whether a string containing a single space contains each character in the sentence (the "word count" in this case is just a count of spaces, so leading or trailing spaces, or extra spaces between words, will be counted as extra words, as your code currently does).
It also allows us to abolish the separate method calls, and the hand-rolled looping code which, as you can see, are prone to small bugs (and I personally would recommend using a For loop rather than Do Until loop in this kind of situation, because For loops are designed to increment the index automatically).
Sub Main()
Console.WriteLine("Sentence Analysis")
Console.WriteLine()
Console.WriteLine("Enter a sentence then press 'Enter'")
Console.WriteLine()
Dim sentence As String
sentence = Console.ReadLine()
Dim vowel_count As Integer = sentence.Count(Function(c) "aeiou".Contains(Char.ToLower(c)))
Dim word_count As Integer = sentence.Count(Function(c) " ".Contains(Char.ToLower(c)))
Console.WriteLine("Your sentence contains {0} words.", word_count)
Console.WriteLine("Your sentence contains {0} vowels.", vowel_count)
Console.ReadLine()
End Sub

Answer 3

You go to the same college as I did. Here is the code...
Option Compare Text
Imports System
Imports System.Diagnostics.Metrics
Imports System.Reflection
Imports System.Text.RegularExpressions
Module Program
Sub Main(args As String())
Console.WriteLine("Sentence Analysis")
Console.WriteLine()
Console.WriteLine("Enter a sentence, then press 'Enter'")
Console.WriteLine()
Dim input As String = Console.ReadLine
Dim letterbyletter() As Char = input.ToCharArray
Dim counter As Integer = 1
Dim vowelcounter As Integer
Call wordcount(input, letterbyletter, counter)
Console.WriteLine("Your sentence contains {0} words.", counter)
Call vowelcount(input, letterbyletter, vowelcounter)
Console.WriteLine("Your sentence contains {0} vowels.", vowelcounter)
End Sub
Sub wordcount(ByVal input As String, ByRef letterbyletter() As Char, ByRef
counter As Integer)
For i = 0 To Len(input) - 1
If letterbyletter(i) = " " Then
counter += 1
End If
Next
End Sub
Sub vowelcount(ByVal input As String, ByRef letterbyletter() As Char, ByRef
vowelcounter As Integer)
Dim vowels() As String = {"a", "e", "i", "o", "u"}
For i = 0 To Len(input) - 1
For j = 0 To 4
If letterbyletter(i) = vowels(j) Then
vowelcounter += 1
End If
Next
Next
End Sub
End Module