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Populate a bidemensional array in VB.Net 2012 (code posted)

Hi,
I am trying to populate an array but I cann't make it work...
I have a 'message' string with the following text (spaces between every word):
"AWVLI QIQVT QOSQO ELGCV IIQWD LCUQE EOENN WWOAO LTDNU QTGAW TSMDO QTLAO QSDCH PQQIQ DQQTQ OOTUD BNIQH BHHTD UTEET FDUEA UMORE SQEQE MLTME TIREC LICAI QATUN QRALT ENEIN RKG"

In my code, firstly, I have to remove spaces from the above text. Then, I have to save the resulting 143 letters into my array called 'nuevo_array' (11rows x 13columns matrix).

The resulting array should contain the following:
AWVLIQIQVTQOS
QOELGCVIIQWDL
CUQEEOENNWWOA
OLTDNUQTGAWTS
MDOQTLAOQSDCH
PQQIQDQQTQOOT
UDBNIQHBHHTDU
TEETFDUEAUMOR
ESQEQEMLTMETI
RECLICAIQATUN
QRALTENEINRKG

    Private Sub btnAnalizar_Click(sender As Object, e As EventArgs) Handles btnAnalizar.Click
        Dim message As String = txtTextoOriginal.Text
        Dim b As String() = message.Split(" ".ToCharArray)

        Dim total_letras As Integer = message.Length '143 letters
        lblLetras.Text = total_letras

        Dim x As Integer = 12
        Dim y As Integer = 10
        Dim nuevo_array(x, y) As String

        For i = 0 To x
            For j = 0 To y
                nuevo_array(x, y) = b.substring(0, 1)
            Next
        Next

    End Sub

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José Perez
Asked:
José Perez
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1 Solution
 
Fernando SotoRetiredCommented:
Hi oscargarin;

Try it like this.

Private Sub btnAnalizar_Click(sender As Object, e As EventArgs) Handles btnAnalizar.Click

     ' Remove all spaces characters from the message
     Dim message As String = txtTextoOriginal.Text.Replace(" ", "")

     Dim total_letras As Integer = message.Length '143 letters
     lblLetras.Text = total_letras

     Dim x As Integer = 12
     Dim y As Integer = 10
     ' Use a char array seeming each character is one array element
     Dim nuevo_array(x, y) As Char
     ' Use to index into the message string
     Dim idx As Integer = 0

     For i = 0 To x
         For j = 0 To y
             nuevo_array(i, j) = message(idx)
             ' Position to next character in the message string
             idx += 1
         Next
     Next

End Sub

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José PerezAuthor Commented:
Sorry I was late.... the answer is perfect!
Thanks a lot.
0
 
Fernando SotoRetiredCommented:
Not a problem, glad I was able to help.
0
 
José PerezAuthor Commented:
:)
0

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