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Finding all the "\n" is a string and removing it

Cyber-EE
Cyber-EE asked
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Last Modified: 2010-04-07
Hello,

I wanted to know how can is there any way to find out if there is "\n" (or more then 1) in my string  and if so remove it.

I tired the Replace function but i wasn't succeesfull

YoniS


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Top Expert 2006

Commented:
u could try using the replace command

newstring = replace(mystring, chr$(13), "")

// sInput = your_string_to_replace
// sCharToRemove = "\n"

Public Function RemoveChar(ByVal sInput As String, ByVal _
   sCharToRemove) As String
Dim iPos As Integer

iPos = InStr(sInput, sCharToRemove)
While iPos > 0
    iPos = InStr(sInput, sCharToRemove)
    If iPos > 0 Then sInput = Left(sInput, iPos - 1) & _
       Mid(sInput, iPos + 1)
Wend

RemoveChar = sInput

End Function
kaliyugkaarjun, there are 5 things wrong with that function, aside from the fact that it can be emulated in one line using the replace function I stated earlier. See if you can find the 4 errors:

Public Function RemoveChar(ByVal sInput As String, _
                                           ByVal strToRemove As String) As String
    Dim iPos As Integer
    iPos = InStr(sInput, strToRemove)
    Do While iPos > 0
        sInput = Left$(sInput, iPos - 1) & Mid$(sInput, iPos + len(strToRemove) )
        iPos = InStr(sInput, strToRemove)
    Loop
    RemoveChar = sInput
End Function

1)One huge error is that the carriage return is often not a single character, but a two character combination, so your function would fail and garble the text as soon as the element to replace was longer than one character, which in this application is big.
2) While/wend is extremely outdated, and no longer has a use, it should not be used
3) Not using the $ character results in more memory consumption and time spent converting the returned Variant back in to a string.
4) You omitted the type declaration for the second String parameter
5) Your loop was malformed, you were resetting iPos at the beginning instead of the end of the loop, which made an if statement necessary when it should not be, and which made one loop occur at the end that did nothing. All this was uneccessary.

Rockiroads, even though you copied my equation and failed to realize that he had allready mentioned attempting the replace function, what you showed has two problems.
1) There is no $ character, again wasting memory and time.
2) chr$(13) is not a universal carriage return and would fail on most operating systems.

I hope I was helpful to everyone including Cyber-EE and the two other commenters.
I appologise if it was not clear, but the function demonstrated above is the correct version of the attempt kaliyugkaarjun made. This is not even necessary however, as the Replace$() function should suffice.

Cyber-EE can you show us the code that you tried, and maybey a part of the string?
I'm pretty sure the solution is up there, he just needs to come back to the thread and read it now.

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