UNIX: Replace Newline w/ Colon, Preserving Newline Before EOF

Posted by Maarx on Stack Overflow See other posts from Stack Overflow or by Maarx
Published on 2010-05-26T18:09:40Z Indexed on 2010/05/26 18:11 UTC
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I have a text file ("INPUT.txt") of the format:

A<LF>
B<LF>
C<LF>
D<LF>
X<LF>
Y<LF>
Z<LF>
<EOF>

which I need to reformat to:

A:B:C:D:X:Y:Z<LF>
<EOF>

I know you can do this with 'sed'. There's a billion google hits for doing this with 'sed'. But I'm trying to emphasis readability, simplicity, and using the correct tool for the correct job. 'sed' is a line editor that consumes and hides newlines. Probably not the right tool for this job!

I think the correct tool for this job would be 'tr'. I can replace all the newlines with colons with the command:

cat INPUT.txt | tr '\n' ':'

There's 99% of my work done. I have a problem, now, though. By replacing all the newlines with colons, I not only get an extraneous colon at the end of the sequence, but I also lose the carriage return at the end of the input. It looks like this:

A:B:C:D:X:Y:Z:<EOF>

Now, I need to remove the colon from the end of the input. However, if I attempt to pass this processed input through 'sed' to remove the final colon (which would now, I think, be a proper use of 'sed'), I find myself with a second problem. The input is no longer terminated by a newline at all! 'sed' fails outright, for all commands, because it never finds the end of the first line of input!

It seems like appending a newline to the end of some input is a very, very common task, and considering I myself was just sorely tempted to write a program to do it in C (which would take about eight lines of code), I can't imagine there's not already a very simple way to do this with the tools already available to you in the Linux kernel.

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