Script to remove only the LAST carriage return and line feed from a text file 1

[Zogfew]

I would like some help with vb script. I know next to nothing about vb scripting so bare with me. I wanted write a dos batch file to do this but quickly realized that this is too complicated for a batch file. After searching the internet I found a vbscript that removes the CR\LF but the only one I found that works, removes all of them. I only want to remove the last one.
This is code I found:

Const ForReading = 1
Const ForWriting = 2

Set objFSO = CreateObject("Scripting.FileSystemObject")
Set objFile = objFSO.OpenTextFile("D:\VDW\VDW_Output\DDOD\TodayDDOD.csv", ForReading)

strText = objFile.ReadAll
objFile.Close
strNewText = Replace(strText, chr(010), chr(013) "") ' chr(010) = line feed chr(013) = carriage return

Set objFile = objFSO.OpenTextFile("D:\VDW\VDW_Output\DDOD\TodayDDOD.csv", ForWriting)
objFile.WriteLine strNewText
objFile.Close

Thanks for the help

 

[guitarzan]

I believe this will work, not tested though:

strNewText = Replace(strText, vbCrLf, "", InStrRev(strText, vbCrLf))

Notes: vbCrLf is a vbscript constant for a chr(10) (carriage return) and a chr(13) line feed

InStrRev will return the position of the search string starting from the end of the main string

http://www.w3schools.com/vbscript/func_instrrev.asp

The 4th parameter of the Replace function is the start position where the replacing will start.

http://www.w3schools.com/vbscript/func_replace.asp

 

[Geates]

do you mean the last instance of a crlf or the last bit of the file contents if it is a crlf?

-Geates

http://raqxtr.sstar.com/caspdoc/html/chapter_5__developer_s_reference.htm

 

[Zogfew]

Geates:
I mean if you sat down and typed up a letter to someone, and you hit the enter key just after the very last word you typed. (if that makes any sense)

Guitarzan:
Itried that code and it removed all of the file contents except for the CR\LF (perhaps I did something wrong?

 

[guitarzan]

My bad, I misread what that 4th parameter does...

 

[Geates]

similar to what guitarzan suggested, remove the last byte if it is a crlf by starting your replace at the last byte.

strNewText = replace(strText, vbCrLf, "", len(strText) - 1)

-Geates

http://raqxtr.sstar.com/caspdoc/html/chapter_5__developer_s_reference.htm

 

[guitarzan]

Sorry for the error earlier (that's why I included the links to the definitions of the functions! ;-))

I think this will do the job

Option Explicit
Const ForReading = 1
Const ForWriting = 2

Dim objFSO, objFile, strText, iPos
Set objFSO = CreateObject("Scripting.FileSystemObject")
Set objFile = objFSO.OpenTextFile("c:\data\vbs\tt1\repl\input.txt", ForReading)

strText = objFile.ReadAll
objFile.Close

iPos = InStrRev(strText, vbCrLf)
If iPos > 0 Then
   strText = Left(strText, iPos - 1) & Replace(Mid(strText, iPos), vbCrLf, "")
   Set objFile = objFSO.OpenTextFile("c:\data\vbs\tt1\repl\input.txt", ForWriting)
   objFile.Write strText
   objFile.Close
End If

 

[guitarzan]

Geates: No, ignore what I said earlier about the Replace function... I was working too quickly and was wrong about what that parameter does.

 

[PHV]

Why not simply this ?

If Right(strText, 2) = vbCrLf Then
  strNewText = Left(strText, Len(strText) - 2)
Else
  strNewText = strText
End If

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[ClulessChris]

strNewText = trim(strText) ??

Never knock on Death's door: ring the bell and run away! Death really hates that!

 

[PHV]

Another way (even simpler) way is to replace this:
objFile.WriteLine strNewText
with thos:
objFile.Write strText

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[guitarzan]

I read the post as remove the last occurrence of "x", from a string, wherever it may occur. Removing a trailing CRLF is a much simpler effort.

 

[Geates]

guitarzan: !! The 4th param: I've always thought it was the position to start looking, not the position to truncate to and then start looking :/

-Geates

http://raqxtr.sstar.com/caspdoc/html/chapter_5__developer_s_reference.htm

 

[Zogfew]

Thanks for the suggestions. I have tried them and have not had luck so far. Perhaps it is me, i'm not sure. Here is a sample of the last three lines of text from the file:

1,112,1,0,15,290,15,16.10,12070,"CK",3,1487,1,194,108.41,0,0,0.00,0,0,5," ",2,0,0.00,0,0,0,0,1,0,0,84.30,0,4,0,0,11513,-3.47,0,0
1,112,1,0,15,292,15,200.00,12915,"WD",4,0,4,15,-287.85,0,0,0.00,0,4,5," ",2,0,0.00,0,0,0,0,1,0,0,-287.85,0,2,0,0,11513,-502.85,0,0
1,112,1,0,15,295,15,15.00,12915,"SC",4,0,4,15,-287.85,0,0,0.00,0,4,5," ",2,0,0.00,0,0,0,0,1,0,0,-287.85,0,2,0,0,11513,-502.85,0,0

See the goofy arrow at the end, it represents the carriage return. When I used Guitarzans code it left the last CR. I tried PHV's
If Right(strText, 2) = vbCrLf Then
strNewText = Left(strText, Len(strText) - 2)
Else
strNewText = strText
End If
with no luck
As far as PHV's post of:
Another way (even simpler) way is to replace this:
objFile.WriteLine strNewText
with thos:
objFile.Write strText

I am unsure which code i should try this in, my original posted code, PHV's, or Guitarzans
Thanks

 

[PHV]

Like this:

Const ForReading = 1 
Const ForWriting = 2 
Set objFSO = CreateObject("Scripting.FileSystemObject") 
Set objFile = objFSO.OpenTextFile("D:\VDW\VDW_Output\DDOD\TodayDDOD.csv", ForReading) 
strText = objFile.ReadAll 
objFile.Close 
Set objFile = objFSO.OpenTextFile("D:\VDW\VDW_Output\DDOD\TodayDDOD.csv", ForWriting) 
objFile.Write strText 
objFile.Close

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[Zogfew]

PHV:
I used the code you posted last and it worked for me.
Thank you for all the help.