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

Question

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.

Answer 1

Sometimes, all you have to do is start to ask the question, and the answer just comes to you. I figured it out while writing my question, but decided to post it anyway. The solution is 'echo'. 'echo' will, also, append a newline. So the solution is:

echo `cat INPUT.txt | tr '\n' ':'` | sed 's/:$//'

Answer 2

This should do the job (cat and echo are unnecessary):

tr '\n' ':' < INPUT.TXT | sed 's/:$/\n/'

Using only sed:

sed -n ':a; $ ! {N;ba}; s/\n/:/g;p' INPUT.TXT

Bash without any externals:

string=($(<INPUT.TXT))
string=${string[@]/%/:}
string=${string//: /:}
string=${string%*:}

Using a loop in sh:

colon=''
while read -r line
do
    string=$string$colon$line
    colon=':'
done < INPUT.TXT

Using AWK:

awk '{a=a colon $0; colon=":"} END {print a}' INPUT.TXT

Or:

awk '{printf colon $0; colon=":"} END {printf "\n" }' INPUT.TXT

Edit:

Here's another way in pure Bash:

string=($(<INPUT.TXT))
saveIFS=$IFS
IFS=':'
newstring="${string[*]}"
IFS=$saveIFS

Answer 3

Here's yet another solution: (assumes a character set where ':' is octal 72, eg ascii)

perl -l72 -pe '$\="\n" if eof' INPUT.TXT