Len() function and spaces

[keyscoob]

I'm using the Len() function to count the number of letters that have been pasted into a Text Box from another application and display the total letter count within a label.

Obviously Len(Text1.Text) will count spaces between words/letters as well as the actual letters. Does anyone now how I can avoid counting the spaces so that my label caption shows only the total number of letters pasted?

Thanks...

[salvadorescobar]

I don' t know what language you are taling about here (is this VBScript?).

You are not wanting to Trim the spaces at the beginning and end, but count all spaces right?

try something like this...

var strCount = Len(Text1.Text);
var strSpaceCount = 0;


// now loop through the string and count the spaces
for (i=0;i<strCount;i++) {
if (Text1.Text[i] == Chr(32)) { // 32 is the ASCII value of a space
strSpaceCount++;
}
}

// finally subtract the no of spaces from the length of the whole string
var strMinusSpacesCount = strCount - strSpaceCount;

There may be a function to count the occurance of chars in a string (I know there is in a number of languages) , however this algorithm will work in many languages.

I hope this helps.

[keyscoob]

Hi,

Thanks for the feedback. I an using VB (should have said) and think I have it sorted now.

Thanks again...

[keyscoob]

Did this to sort my problem:

Private Sub Command1_Click()

'Label1.Caption = Len(Text1.Text)

Dim strToSearch As String
Dim intLength As Integer
Dim intPosition As Integer
Dim intCount As Integer
Dim i As Integer
Dim j As Integer

Dim arrTotals(26) As Integer
Dim arrPercentages(26) As Single
Dim strMessage As String

'put the text from the text box into a string
strToSearch = Text1.Text

'get the string length
intLength = Len(strToSearch)

'only make checks if it is not an empty string
'otherwise you will get an overflow error
If intLength > 0 Then

For i = 64 To 90
'for each letter
intCount = 0
Do
'check the number of occurances
intPosition = InStr(intPosition + 1, UCase(strToSearch), Chr(i))
If intPosition > 0 Then
intCount = intCount + 1
Else
Exit Do
End If
Loop
arrTotals(i - 64) = intCount
arrPercentages(i - 64) = Format((intCount / intLength) * 100, "0.00")
Next i

'display results
strMessage = ""

For i = 1 To 26
strMessage = strMessage & Chr(i + 64) & " " & "Appears " & arrTotals(i) & " Times" & " And Accounts For " & arrPercentages(i) & "% Of The Total Characters" & ", "
Next i

End If

Text2.Text = strMessage

End Sub


Another Question!!

My textbox has 26 sets of results displayed within it by using:

For i = 1 To 26
strMessage = strMessage & Chr(i + 64) & " " & "Appears " & arrTotals(i) & " Times" & " And Accounts For " & arrPercentages(i) & "% Of The Total Characters" & ", "
Next i

I have the TextBox set to multiline with scroll bars. How do I get the results for each loop of the above For-Next to be displayed on a new line within the text box????

[Plasma]

When you add each result to the Text box also do a & vbNewLine

eg....

Text1.Text = Result1 & vbNewLine

Hope this helps, let me know if not!

Plasma

[keyscoob]

Hi,m8

Thanks for the feedback. In the end I used the vbCrLf function in the following format.

Text2.Text = strMessage + vbCrLf & "The total number of Letters is " & intLength _
& " (excluding spaces)" & vbCrLf

Many thanks for your feedback the app now works virtually how I need it to.

[graham.edmon]

why not just count the spaces and subtract it from the length.. it is slow an unnecessary to go through every character of the string...

something like this

Dim lSpaces
Dim lIndex

lSpaces = 0
lIndex = Instr(1, sString, " ")
Do Until lIndex = 0
lSpaces = lSpaces + 1
lIndex = Instr(lIndex, sString, " ")
Loop

Msgbox "Chars is string are " & Len(sString) - lSpaces

[RhinoBanga]

it is slow an unnecessary to go through every character of the string

Mind you the Instr function is going thru the list as well ... albeit more efficiently than VB code.