Parsing the colors

[Networks]

I need a little help trying to parse messages that have been tweaked so they allow your messages to be different colors. ( Trivia Bot does, Stealthbot can parse it. )

I am sorry I seem so newbish right, It's morning and I can't remember what this is called again.

I am not asking for an entire function just a little help. * Pretty sure you'll understand *

[Noodlez]

Colors from which client?

[GoSuGaMING]

hes talking about stracraft when you do ÁU the color displays Blue on the client he wants it where if the bot see's that then the "addc" for that text will  be blue

[Newby]

Split by Á

Then for each item in the array you split the message by, determine the color, and add the rest of that array item to the RTB.

[Networks]

Split by Á

Then for each item in the array you split the message by, determine the color, and add the rest of that array item to the RTB.

I understand arrays and such but how do I split all the word(s) with Á in front and add it to the array? Every single one and then have it also change it to actual colors in the Richtextbox?

[Fr0z3N]

Code Select Expand

if (message.indexOf("Á") > 0) {
  String array[] = null;
   array = message.split("Á");
{

if instr(message, "Á") then
dim array() as string
array = split(message, "Á")
end if

Edit: Code tags.

[Eli_1]

Just an extension to what fr0zen already posted.
NOTE: This code was untested and was found to not work. I've fixed the code and it can be found here.

Code Select Expand


'// Put me somewhere  :P
Const Message as String = "ÁUThis is an example string. ÁYBlah blah blah blah blah (more useless text). ÁRInsert some retarded comment here. ÁVBlah blah blah blah blah." & vbCrLf
'/* This example will Split this string. */

Dim ColorParse() as String
'/* MerF!? */

Dim ColorByte as String
'/* This is just to make the code a little
'** easier to read. */

If InStr(1, Message, "Á", vbTextCompare) <> 0 Then
   ColorParse = Split(Message, "Á")
   For i = 0 to UBount(ColorParse)
       ColorByte = Left(ColorParse(i), 1)
       ColorParse(i) = Right(ColorParse(i), Len(ColorParse(i) - 1)
       AddText txtOutput, GetColor(ColorByte), ColorParse(i)
   Next i
End If




GetColor()

Code Select Expand


Public Function GetColor(byval ColorByte as String) as vbColorConstants
'// I think it's vbColorConstant. It might be different :P
Select Case ColorByte
. . .
End Select
. . .
End Function


[Networks]

Thank you Eli_1

[Networks]

Code Select Expand


Dim ColorParse() As String
'/* MerF!? */

Dim ColorByte As String
Dim i As Integer
'/* This is just to make the code a little
'** easier to read. */

If InStr(1, Message, "Á", vbTextCompare)  0 Then
   ColorParse = Split(Message, "Á")
   For i = 0 To UBound(ColorParse)
       ColorByte = Left(ColorParse(i), 1)
       ColorParse(i) = Right(ColorParse(i), Len(ColorParse(i) - 1))
       'AddText txtOutput, GetColor(ColorByte), ColorParse(i)
   Next i
End If


its says

variable required - can't assign to this expression
at:
ColorParse(i) = Right(ColorParse(i), Len(ColorParse(i) - 1))

Why?

[Eli_1]

at:
ColorParse(i) = Right(ColorParse(i), Len(ColorParse(i) - 1))

Ya, that's my bad. I wrote that code on the fly when I was posting and never tested it. That's happening because of a mis-placed parenthasis.

Fixed.

Code Select Expand


ColorParse(i) = Right(ColorParse(i), Len(ColorParse(i)) - 1)


[BaDDBLooD]

I was looking at this, and i placed a message box after:

Code Select Expand



   If InStr(1, Message, "Á", vbTextCompare) = 1 Then

[/vbcode]

and friend said something with Á message box didn't go off

[FuzZ]

I was looking at this, and i placed a message box after:

Code Select Expand



   If InStr(1, Message, "Á", vbTextCompare) = 1 Then

[/vbcode]

and friend said something with Á message box didn't go off

Code Select Expand


If InStr(Message, "Á", vbTextCompare) >= 1 Then
    MsgBox "Found Á"
end if

If InStr(Message, "Á", vbTextCompare) <> 0 Then
   MsgBox "Found Á"
end if


Edit> added the latter.

[Eli_1]

Code Select Expand


If InStr(Message, "Á", vbTextCompare) >= 1 Then
   MsgBox "Found Á"e
end if

That's not the syntax InStr takes.
It should be InStr(Start, S1, S2, Seach type).
Your doing InStr(S1, S2, Search type).

This should work:

Code Select Expand


InStr(1, Message, "Á", vbTextCompare) >= 1 Then
   MsgBox "Found Á"e
end if



And to BaDDBLooD,
If InStr(1, Message, "Á", vbTextCompare) = 1 Then.
InStr returns the first instance of the "search string" (Á in this instance). That's why you do <> 0. InStr can return any number greater than 0, and less than or equal to Len(Message) -- depending on where in the string Á is.

[BaDDBLooD]

yeah it still doesn't work

EDIT: Actually... the InStr function has NEVER worked for me... ;[

[Eli_1]

If InStr(1, Message, "Á", vbTextCompare) returns 0, then 'Message' doesn't contain that character...

And I just tryed testing the code I wrote, and there's more than 1 error. I'm fixing it right now -- I'll post when I finish.