Coding help with export

[Mike Schwartz]

     Within your processing, you can always add chr("10")'s and chr("13")'s
yourself at the end of the line in any combination that you want.  Then you
don't have to worry about /R's and /N's or any extracurricular conversion
scripts.

 

Mike Schwartz

 

  _____  

From: filepro-list-bounces+mschw=athenet.net at lists.celestial.com
[mailto:filepro-list-bounces+mschw=athenet.net at lists.celestial.com] On
Behalf Of Del
Sent: Thursday, April 12, 2007 1:49 PM
To: filePro Mailing List
Subject: Coding help with export

 

Hi:

 

I feel dumb asking this, but I have wasted a lot of time already without a
solution, so maybe it is not as easy as I thought it should be, either that
or my brain is just not with it today (if so, not the first time).

 

Under SCO Unix, I am using the "export word" command to produce a flat ASCII
file.  Export word defaults to a record (or line) delimiter of \n (newline).
I want to change this to \r\n (carriage return/newline) for compatibility
with some windows software that the data is passed to.  Well, I have not
been able to find any way to do this with "export word" or even with "export
ascii" where I thought I could use r=\r\n, but this gives me a syntax error.

 

I even tried to issue a SED substitution command after generating the output
file:

 

sed s/\n/\r\n/ filename >target filename

sed s/"\n"/"\r\n"/ filename >target filename

 

I also tried:

 

tr "\n" "\r\n"  - this changed the \n to \r but not to \r\n.

 

None of this worked, I really don't know much about sed  or tr and regular
expressions, but thought this had a shot.

 

So is there any way to export an ASCII file and generate a record delimiter
of carriage return/linefeed? Or convert it after generating it with standard
Unix commands?

 

Del

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