Ubuntu – How to replace multiple lines with single word in file(inplace replace)

awkcommand lineperlsedtext processing

Content of my filename file is as following(for example):

My block of line starts from here 
START
First line
second line
third line
END
and end to here for example.

I want to replace block of lines between START and END with just a single word, for example with SINGLEWORD. Like below:

My block of line starts from here 
SINGLEWORD
and end to here for example.

I can find my block of lines with using this command:

grep -Pzo "START(.|\n)*END" filename

And the result of running above command will be like this:

START
First line
second line
third line
END

Then I used this command to combine all lines into a single line:

LAST_RESULT | sed -e :a -e '/$/N; s/\n/ /; ta'

Then I will get this result:

START First line second line third line END

And with my last command LAST_RESULTS | sed 's/.*/SINGLEWORD/' I change them to "SINGLEWORD" and I get this result.

SINGLEWORD

Now what I want is: How can I use this command(Or your suggestion command) and replace(in place)my block of lines to "SINGLEWORD" word? And the final result will be like this file:

My block of line starts from here 
SINGLEWORD
and end to here for example.

Best Answer

This can be done very easily in perl:

$ perl -i -p0e 's/START.*?END/SINGLEWORD/s' file
$ cat file
My block of line starts from here 
SINGLEWORD
and end to here for example.

Explanation

-0 sets the line separator to null

-p apply the script given by -e to each line and print that line

The regexp modifier:

  • /s Treat string as single line. That is, change . to match any character whatsoever, even a newline, which normally it would not match.

Why the ?:

  • By default, a quantified subpattern is "greedy", that is, it will match as many times as possible (given a particular starting location) while still allowing the rest of the pattern to match. If you want it to match the minimum number of times possible, follow the quantifier with a ?.
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