bash script to display tail

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# 1  
Old 04-08-2011
bash script to display tail

Hi everyone, I'm trying to write a script to format a file using unix2dos. I want to output all but the first 14 lines in a file. Then I want to pipe this to unix2dos to convert the output to a file that's easily readable on windows. Here's what I have:
export Lines=`wc -l < $1`
export Lines=`expr $Lines - 14`
tail -n $Lines $1 | unix2dos print.txt

It keeps giving me errors though. Any thoughts?
# 2  
Old 04-08-2011
Gives you what errors?

---------- Post updated at 04:03 PM ---------- Previous update was at 04:00 PM ----------

Anyway, after checking some manpages, depending on what version of tail you have you might be able to do this:

tail -n +15 < inputfile | sed 's/$/\r//' > winfile

+15 as in, start printing at line 15, ignoring lines 1-14..

Don't even need unix2dos.
# 3  
Old 04-08-2011
What does the whole " | sed 's/$/\r//' > winfile" part do?
# 4  
Old 04-08-2011
It removes carriage returns (\r (^M)) from the end of lines.

Unix text files are usually very easily readable on Windows, if you change from Notepad as your default application for viewing text files.
# 5  
Old 04-08-2011
I just tried that, and it reports:

sed: -e expression #1, char 8: unknown option to `s'
: command not found
# 6  
Old 04-08-2011
Originally Posted by LuminalZero
I just tried that, and it reports:

sed: -e expression #1, char 8: unknown option to `s'
: command not found
Post exactly what you tried (using code tags ).
# 7  
Old 04-08-2011
I made the script file

tail -n +15 < $1 | sed 's/$/\r//' > winfile

Then on the command line I typed:
./ test.cpp

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