Way to remove blank lines from a single string?

This shouldn’t be all that hard, but i’ve spent much of the afternoon on what should be something very simple: removing the trailing and leading carriage returns from a single text string.

given something like:

"text here

"

I’ve tried trim, trim$ full trim, converting my string to a Cstr, i even tried Replace with a Chr(10). Nothing worked, it always returns the extra lines

Sub Click(Source As Button)

Dim workspace As New NotesUIWorkspace

Dim uidoc As NotesUIDocument

Dim update As String



Set uidoc = workspace.CurrentDocument

update= uidoc.FieldGetText( "NewUpdate" )

'update = Replace(update, "", "____")

'update = Fulltrim(Cstr(update))

update = Trim(Cstr(update))

update = Trim$(Cstr(update))

Messagebox "The Update is:" &Chr(10)& Trim(update)

Messagebox "The Update is:" &Chr(10)& Trim$(update)

Messagebox "The Update is:" &Chr(10)& Fulltrim(update)

End Sub

Subject: Way to remove blank lines from a single string?

You can try this.

Public Function ReplaceSubstring(Source As String, Find As String, ReplaceStr As String ) As String 

	Dim findLen As Long 

	Dim replaceLen As Long 

	Dim index As Long 

	Dim counter As Long 

	

	findLen = Len(Find) 

	replaceLen = Len(ReplaceStr) 

	

	If findLen > 0 Then 

		index = 1

		ReplaceSubstring = Source 

		Do 

			index = Instr(index, ReplaceSubstring, Find, 5) 

			If index = 0 Then Exit Do 

			If findLen = replaceLen Then 

				Mid$(ReplaceSubstring, index, findLen) = ReplaceStr 

			Else 

				ReplaceSubstring = Left$(ReplaceSubstring, index - 1) & ReplaceStr & Mid$(ReplaceSubstring, index + findLen) 

			End If 

			index = index + replaceLen 

			counter = counter + 1 

		Loop 			

		If Start > 1 Then ReplaceSubstring = Mid$(ReplaceSubstring, Start) 

	Else

		ReplaceSubstring = Source

	End If

	

End Function 



Function ClearString(Source As String) As String

	ClearString = Source		

	ClearString = ReplaceSubString( ClearString, Chr(13), "" )

	ClearString = ReplaceSubString( ClearString, Chr(10), "" )

End Function

Subject: I think the Chr(13) did the trick

I was using Chr(10)

update = Replace(update, Chr(13), “”)

update = Fulltrim(update)

It seems i still need both statements, though i;m not quite sure why, either way it looks like it’s good enough!

Thanks!

Subject: *Internally, Notes stores carriage returns as Chr(0), not Chr(13) or Chr(10)

Subject: *Internally, Notes stores carriage returns …

But for the code actually running in the client, isn’t it still the platform’s convention, that has to be obeyed, not the internal storage format?

This works (without the need for a further fulltrim)

update = Replace(update, Split(Chr(13) & “,” & Chr(10), “,”), “”)

but this does not

update = Replace(update, Chr(0), “”)