How can I replace a newline with sed?

Question

I've got text with whitespace (including a newline) that I need to replace. I'd like to use sed, but AIX doesn't seem to support the answers to sed newline questions I found. The version of bash I'm using is "GNU bash, version 3.00.16(1)-release (powerpc-ibm-aix5.1)"

If I run a command that outputs a bunch of pairs of lines of text like:

alias:  aliasname
                10:00:00:00:00:00:00:00

What is the best way to make it one line for each? If I use ssh user@system command | tr '\n' ' ', it replaces both newlines, and I need to keep the last one or else the output goes into a single line.

Edit: what I've tried so far is:

• | sed -e ':1' -e 'N' -e '$!b1' -e 's/\n/ /g' which replaces all the newlines, putting the entire output stream into a single line.
• | while read i; do printf "%s " $i; done; echo which also replaces all the newlines
• | tr -d '\n' and | tr -d '\n'; echo, which still replace all newlines, however the echo adds one back at the last command.

Answer 1

Easyest

ssh user@system command | tr '\n' ' ';echo

Or if you want sed

ssh user@system command | sed 'N;s/\n\s\+/ /'

Answer 2

Try this:

sed -e ':1' -e 'N' -e '$!b1' -e 's/\n/ /g'

Answer 3

Here's a pure bash approach that will delete the first newline of every pair:

command | 
  while IFS=$'\n' read i; do 
    let c++; [ "$(expr $c % 2)" -eq "0" ] && echo "$i" || printf "%s " "$i"; 
  done

If you don't need to keep the whitespace unchanged, you can leave out the IFS:

command | 
  while read i; do 
    let c++; [ "$(expr $c % 2)" -eq "0" ] && echo "$i" || printf "%s " "$i"; 
  done

The same idea using perl:

command | perl -ne 'chomp;$. % 2 == 0 ? print "$_\n" : print "$_"' 

Answer 4

It probably best to do this with sed; see Why is using a shell loop to process text considered bad practice?  For example, this minor variation of Costas’s second (sed) answer:

sed 'N; s/\n/ /'

does a pretty good job of answering the question: N to append the next line of input into the pattern space, and then s/\n/ / to do the substitution.  (And then these commands repeat for every two lines of input.)

But, if you want to do it entirely in the shell, this variation of terdon’s answer:

while IFS= read -r line1
do
    IFS= read -r line2; printf '%s %s\n' "$line1" "$line2"
done

works without running any external program(s).  The logic is (IMHO) simple: while not at the end of the input, read a line, and then read another line, and then output the two strings, separated by spaces, on one line.

The above code has a problem with an edge case: if the input is an even number of well-formed lines followed by some characters not terminated with a newline, it discards the final (partial) line.  (terdon’s answer has the same issue, but not just for even numbers of lines.)  This is because read returns failure when it reads a partial line, even though it does put the data into the variable.  So the fix is this:

while IFS= read -r line1  ||  [ "$line1" != "" ]
do
    IFS= read -r line2; printf '%s %s\n' "$line1" "$line2"
done

which keeps going until read fails and sets the variable to an empty string.