Find the length of a word in an array - Visual Basic - vb.net

Question

I have a richtextbox and I converted the words to an array, then I have code that will take the length and the output it... only problem is I don't know how to loop through all the different values of the array to check the length of each individual part of the array.
(I have this set on TextChanged)
Dim len1, len2, len3, len4, len5, len6, len7, len8, len9, len10, len11, len12, len13, len14, len15 As Integer
Dim input As String
Dim words As String()
Dim length As Integer
input = RichTextBox1.Text
words = input.Split(New String() {" "}, StringSplitOptions.RemoveEmptyEntries)
For Each w In words
length = Len(w)
Select Case length
Case 1
len1 = len1 + 1
Case 2
len2 = len2 + 1
Case 3
len3 = len3 + 1
Case 4
len4 = len4 + 1
Case 5
len5 = len5 + 1
Case 6
len6 = len6 + 1
Case 7
len7 = len7 + 1
Case 8
len8 = len8 + 1
Case 9
len9 = len9 + 1
Case 10
len10 = len10 + 1
Case 11
len11 = len11 + 1
Case 12
len12 = len12 + 1
Case 13
len13 = len13 + 1
Case 14
len14 = len14 + 1
Case 15
len15 = len15 + 1
End Select
Next
letcount.onelet.Text = Val(len1)
letcount.twolet.Text = Val(len2)
letcount.threelet.Text = Val(len3)
letcount.fourlet.Text = Val(len4)
letcount.fivelet.Text = Val(len5)
letcount.sixlet.Text = Val(len6)
letcount.sevenlet.Text = Val(len7)
letcount.eightlet.Text = Val(len8)
letcount.ninelet.Text = Val(len9)
letcount.tenlet.Text = Val(len10)
letcount.elevenlet.Text = Val(len11)
letcount.twelevelet.Text = Val(len12)
letcount.thirteenlet.Text = Val(len13)
letcount.fourteenlet.Text = Val(len14)
letcount.fifteenlet.Text = Val(len15)

Answer 1

bendataclear showed you how to use a For Next loop. If you can use LINQ, you can use the following code:
Sub Main
Dim words As String() = {"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten" }
Dim lengthsOnly = words.Select(Function(w) w.Length).ToArray()
Dim wordsAndLengths = words.Select(Function(w) New With { .Word = w, .Length = w.Length }).ToArray()
End Sub
The first Select gets only the word lengths, the second returns an array with each word and it's length.

Answer 2

David,
Some clarification for you:
w is a string. The compiler determines the type at design time. See this S.O. thread
What does VB.Net For Each Loop look at to Infer the Type
Dim words() As String = {"0", "1", "2"}
For Each w In words
System.Windows.Forms.MessageBox.Show(w)
Next
The type is inferred by the compiler by looking at the elements in the IEnumerable collection (array, list, etc.) The intellisense even knows it's a string at design time, even when w looks like it is defined as a generic. Try it yourself...
BUT. You can of course add the type explicitly
Dim words() As String = {"0", "1", "2"}
For Each w As String In words
System.Windows.Forms.MessageBox.Show(w)
Next
The string array example doesn't really demonstrate the convenience of the shorthand. See below
Dim dictionaryOfDictionaries As New Dictionary(Of String, Dictionary(Of String, String))
For Each d As KeyValuePair(Of String, Dictionary(Of String, String)) In dictionaryOfDictionaries
' do something with each dictionary d
Next
' is identical to
For Each d In dictionaryOfDictionaries
' do something with each dictionary d
Next

Answer 3

You can ignore double spaces and enters because they are considered empty:
'HOW CAN I EXCLUDE DOUBLE SPACES
'AND INCLUDE IF SOMEBODY PRESSES ENTER?
words = input.Split(New String() {" "}, StringSplitOptions.RemoveEmptyEntries)

Answer 4

You could use another array?
Dim arr(14) as integer
Dim input As String
Dim words As String()
Dim length As Integer
input = RichTextBox1.Text
words = Split(input, " ")
For Each w in words
Dim l as Integer = Math.Min(Len(w) - 1,14)
arr(l) = arr(l) + 1
Next
Edit
In answer to questions in the comment:
w is a variable, and it is declared in the shortcut code For Each w in words
The shortcut declares a variable w then populates it with each member of the array words and runs the code between the For and next.
A longer way to do this without the shortcut would be something like:
Dim i as Integer
For i = 0 to words.GetUpperBound(0) -1
Dim w as string
w = words(i)
i = i + 1
'Rest of code
Next i
The next part Dim l as Integer = Math.Min(Len(w) - 1,14) sets l to the length of the word up to 14, as if we have a word longer than 15 letters it won't fit in the array.
Another way to do it would be:
Dim l as Integer
l = Len(w) - 1
If l > 14 then
l = 14
End If
arr(l) = arr(l) + 1