SED - removing string followed by LineFeed (\n)

Question

I can't find a suitable sed expression to remove a word followed by a line return (\n)

Test file is:

line1\n
line2\n
line3mark\n
line4\n
line5\n

and i want to remove all occurances of mark\n leaving, in this case:

line1\n
line2\n
line3line4\n
line5\n

have searched and can use:

sed 's/\n//g' test.file  to remove ALL \n's

but

sed 's/mark\n//g' test.file does not work

Strangely, s/mark\n//g does seem to work ok in vi in interactive mode.

Any help greatly appreciated! I would like to understand how to do it using SED if possible as I am sure it is possible!! However, if it can be done another way then I'm also happy as long as its on the command line as it has to run over many files.

Many thanks.

Answer 1

This should do the trick:

sed -i ':a;N;$!ba;s/mark\n//g' file

Explanation:

;    command separator within sed
:a   a label, like in C/C++
N    appends the next line to the pattern space
$!ba repeats the N command for all lines but the last line

sed proceeds like this. it reads the standard input into the pattern space, performs a sequence of editing commands on the pattern space, then writes the pattern space to STDOUT.

When you do something like

sed -i 's/mark\n//' file

lines are copied to the pattern space one by one.

:a;N;$!ba appends each line to the pattern space.

Then the pattern space can be processed in one pass, removing any mark\n , the g option, for global, is important here because it ask sed not to stop at the first matching pattern.