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Avoid carriage return until ^M is found (CentOS 6, bash 4.1)

Hi everyone,
I have the following contents in a text file (as seen when viewed using vim):

Code:

one two three ^M

four five six ^M

seven

eight

nine ^M

ten eleven twelve ^M

(That is just a small portion of the file)
How can I obtain the following result?

Code:

one two three ^M

four five six ^M

seven eight nine ^M

ten eleven twelve ^M

If there are words in a certain line that does not end with ^M, they should be all in one line until a ^M appears.
Thanks in advance.

Give this a try:

Code:

sed -i ':loop;/^M/!{N;bloop};s/\n/ /g' infile

The ^M is special and can be created like this: type ctrl-v then ctrl-m

As a side effect the ^M's are not longer visible when opened in vi. This because the file is now a dos file and not a mixed dos/linux file.

Below a dump of the input file.

Before:

Code:

$ od -c infile

0000000  o  n  e      t  w  o      t  h  r  e  e      \r  \n

0000020  f  o  u  r      f  i  v  e      s  i  x      \r  \n

0000040  s  e  v  e  n  \n  e  i  g  h  t  \n  n  i  n  e

0000060      \r  \n  t  e  n      e  l  e  v  e  n      t  w

0000100  e  l  v  e      \r  \n

0000107

After:

Code:

0000000  o  n  e      t  w  o      t  h  r  e  e      \r  \n

0000020  f  o  u  r      f  i  v  e      s  i  x      \r  \n

0000040  s  e  v  e  n      e  i  g  h  t      n  i  n  e

0000060      \r  \n  t  e  n      e  l  e  v  e  n      t  w

0000100  e  l  v  e      \r  \n

0000107

The \r \n sequence represents the dos newline (the ^M). A single \n is a linux newline.

You could also try awk:

Code:

awk -F"[ \n]" 'BEGIN{ORS=RS="\r\n"}$1=$1' file