String Manipulation

[cnicolapc]

Hi,
does anyone know how to convert a string that contains CRLF in a string of text without CRLF?
Thank You.
Nicola

 

[NJDude]

Something like this:

Dim test, new As String
            
test = "hello" & CRLF & "there"
            
Msgbox(test, "")
            
new = test.Replace(CRLF, " ")
            
Msgbox(new, "")

 

[cnicolapc]

hi,
Thank you very much.
I tried to solve this problem, in my listview looks like.
Have an idea?
Thank You
Nicola

 

[NJDude]

Maybe the string has only LF or CR and not both.

 

[rbsoft]

Depends where you text with CRLF was created.

CRLF in VB or PB (Windows) is Chr(13) & Chr(10) (CarriageReturn + LineFeed).
In Android it is only Chr(10). So you might have to remove Chr(13) still.

See this link:
http://www.b4x.com/forum/basic4andr...-up-rectangles-multiline-label.html#post60471

Rolf

 

[cnicolapc]

Hi Rolf,
Thank you
I tried but the result remains the same.
Text is hidden in Listview as in the previous post.
This is a part of code that I inserted.

Valore2=Cursor1.getstring("Cosa")
      Valore2=Valore2.Replace(Chr(13),"")
            ListaCose.AddTwoLines2(Valore2,Valore3,Valore)
            Next

I also tried with
Valore2=Valore2.Replace(Chr(10),"")
but with no result

Thank You
Nicola

 

[rbsoft]

Have you tried...

Valore2=Valore2.Replace(Chr(13) & Chr(10),"")

Rolf

 

[cnicolapc]

Hi Rolf,
Thanks for your help, but the problem is not solved.
When I put the text I press Enter on the keyboard (so the return should Chr (13) & Chr (10) )
or perhaps it could be a different code?
Nicola

 

[klaus]

Could you post your code with the text you want to display in the ListBox ?
Or at least a small example project.
Otherwise it's difficult to help you.

Best regards.

 

[cnicolapc]

Hi Klaus
here is the code

For i = 0 To Cursor1.RowCount - 1
        Cursor1.Position = i
      valore=Cursor1.GetString("Codice")
      valore2=Cursor1.getstring("Cosa")
      Valore3=Cursor1.getstring("Dove")
      If valore2.Length >=30 Then
      Valore2=valore2.SubString2(0,30)&"..."
      End If
      If valore3.Length >=39 Then
      Valore3=valore3.SubString2(0,39)&"..."
      End If
      
            ListaCose.AddTwoLines2(Valore2.Replace(Chr(13)& Chr(10),""),Valore3,Valore)
            Next
   Cursor1.Close

Thank You
Nicola

 

[klaus]

What exactly is the content of Valore2 ?

In the meantime try this code:

For i = 0 To Cursor1.RowCount - 1
  Cursor1.Position = i 
  Valore=Cursor1.GetString("Codice")
  Valore2=Cursor1.getstring("Cosa")
  Valore2 = Valore2.Replace(Chr(13),"")
  Valore2 = Valore2.Replace(Chr(10),"")
  Valore3=Cursor1.getstring("Dove")
  If Valore2.Length >=30 Then
    Valore2=Valore2.SubString2(0,30)&"..."
  End If         
  If Valore3.Length >=39 Then
    Valore3=valore3.SubString2(0,39)&"..." 
  End If                          
   ListaCose.AddTwoLines2(Valore2,Valore3,Valore)
Next     
Cursor1.Close

Best regards.

 

[vb1992]

You can see what is going on a little with this too

Sub Activity_Create(FirstTime As Boolean)

Dim theline As String

theline = Chr(13) & "We love to go to " & CRLF & " the store." & Chr(9) & Chr(13)

cleanword(theline)

End Sub


Sub cleanword(bigword As String)

Dim c As Int
Dim BuildWord As String
'
' go through the bigword - character by character
'
Log("##################### START ##################################")
Log("Big Word Length is: " &  bigword.Length )

For c = 0 To bigword.Length

   ' Examine each character that is not
   ' a valid character and report it to the log
   'BuildWord = BuildWord & bigword.CharAt(c)
   'Log(BuildWord)
   
   MyASci = Asc(bigword.CharAt(c)) 

 
   
      If MyASci < 32 Then Log("**Chr(" & MyASci & ") was found! at Position: " & c )
      If MyASci > 126 Then Log("**Chr(" & MyASci & ") was found! at Position: " & c )
       
      
 
Next

End Sub

 

[rbsoft]

This line

For c = 0 To bigword.Length

needs to be changed to

For c = 0 To bigword.Length - 1

Otherwise you will get an error

That code just shows that CRLF adds a Chr(10). Did you examine Valore2 in your original program with your cleanword routine?

Rolf

 

[rbsoft]

it seems the problems is in the Replace function - it does not replace chr(10).

Attached is a little test project that shows the problem.

 

[klaus]

We are making an error:

t2.Replace(Chr(13)," ")
t2.Replace(Chr(10)," ")

must be

[COLOR=Red]t2 =[/COLOR] t2.Replace(Chr(13)," ")
[COLOR=Red]t2 =[/COLOR] t2.Replace(Chr(10)," ")

Best regards.

 

[agraham]

Strings are immutable in Java (and C#) so String functions return a new String and do not alter the existing one. You will have to assign the returned String to the existing variable if you want the effect of altering a String.

 

[rbsoft]

:BangHead:
Manchmal sieht man den Wald vor lauter Bäumen nicht!

Sometimes you don't see the forest because of all the trees!
(Germam proverb)

Thanks Klaus and Andrew

Rolf

 

[cnicolapc]

Hi,
Thank you all for taking the time to support this forum.
Thanks
Nicola

 

[cnicolapc]

Hi Rolf,
You managed to get the correct result?
Nicola

 

[rbsoft]

Yes, works now.
I uploaded the corrected project.

(See Klaus' post.)

Rolf