Replace entire paragraph with another from linux command line - c++

Question

The problem I have is pretty straightforward (or so it seems). All I want to do is replace a paragraph of text (it's a header comment) with another paragraph. This will need to happen across a diverse number of files in a directory hierarchy (source code tree).
The paragraph to be replaced must be matched in it's entirety as there are similar text blocks in existence.
e.g.
To Replace
// ----------
// header
// comment
// to be replaced
// ----------
With
// **********
// some replacement
// text
// that could have any
// format
// **********
I have looked at using sed and from what I can tell the most number of lines that it can work on is 2 (with the N command).
My question is: what is the way to do this from the linux command line?
EDIT:
Solution obtained: Best solution was Ikegami's, fully command line and best fit for what I wanted to do.
My final solution required some tweaking; the input data contained a lot of special characters as did the replace data. To deal with this the data needs to be pre processed to insert appropriate \n's and escape characters. The end product is a shell script that takes 3 arguments; File containing text to search for, File containing text to replace with and a folder to recursively parse for files with .cc and .h extension. It's fairly easy to customise from here.
SCRIPT:
#!/bin/bash
if [ -z $1 ]; then
echo 'First parameter is a path to a file that contains the excerpt to be replaced, this must be supplied'
exit 1
fi
if [ -z $2 ]; then
echo 'Second parameter is a path to a file contaiing the text to replace with, this must be supplied'
exit 1
fi
if [ -z $3 ]; then
echo 'Third parameter is the path to the folder to recursively parse and replace in'
exit 1
fi
sed 's!\([]()|\*\$\/&[]\)!\\\1!g' $1 > temp.out
sed ':a;N;$!ba;s/\n/\\n/g' temp.out > final.out
searchString=`cat final.out`
sed 's!\([]|\[]\)!\\\1!g' $2 > replace.out
replaceString=`cat replace.out`
find $3 -regex ".*\.\(cc\|h\)" -execdir perl -i -0777pe "s{$searchString}{$replaceString}" {} +

Answer 1

find -name '*.pm' -exec perl -i~ -0777pe'
s{// ----------\n// header\n// comment\n// to be replaced\n// ----------\n}
{// **********\n// some replacement\n// text\n// that could have any\n// format\n// **********\n};
' {} +

Answer 2

Using perl:
#!/usr/bin/env perl
# script.pl
use strict;
use warnings;
use Inline::Files;
my $lines = join '', <STDIN>; # read stdin
my $repl = join '', <REPL>; # read replacement
my $src = join '', <SRC>; # read source
chomp $repl; # remove trailing \n from $repl
chomp $src; # id. for $src
$lines =~ s#$src#$repl#gm; # global multiline replace
print $lines; # print output
__SRC__
// ----------
// header
// comment
// to be replaced
// ----------
__REPL__
// **********
// some replacement
// text
// that could have any
// format
// **********
Usage: ./script.pl < yourfile.cpp > output.cpp
Requirements: Inline::Files (install from cpan)
Tested on: perl v5.12.4, Linux _ 3.0.0-12-generic #20-Ubuntu SMP Fri Oct 7 14:56:25 UTC 2011 x86_64 x86_64 x86_64 GNU/Linux

Answer 3

This might work:
# cat <<! | sed ':a;N;s/this\nand\nthis\n/something\nelse\n/;ba'
> a
> b
> c
> this
> and
> this
> d
> e
> this
> not
> this
> f
> g
> !
a
b
c
something
else
d
e
this
not
this
f
g
The trick is to slurp everything into the pattern space using the N and the loop :a;...;ba
This is probably more efficient:
sed '1{h;d};H;$!d;x;s/this\nand\nthis\n/something\nelse\n/g;p;d'
A more general purpose solution may use files for match and substitute data like so:
match=$(sed ':a;N;${s/\n/\\n/g};ba;' match_file)
substitute=$(sed ':a;N;${s/\n/\\n/g};ba;' substitute_file)
sed '1{h;d};H;$!d;x;s/'"$match"'/'"$substitute"'/g;p;d' source_file
Another way (probably less efficient) but cleaner looking:
sed -s '$s/$/\n###/' match_file substitute_file |
sed -r '1{h;d};H;${x;:a;s/^((.*)###\n(.*)###\n(.*))\2/\1\3/;ta;s/(.*###\n){2}//;p};d' - source_file
The last uses the GNU sed --separate option to treat each file as a separate entity. The second sed command uses a loop for the substitute to obviate .* greediness.

Answer 4

As long as the header comments are delimited uniquely (i.e., no other header comment starts with // ----------), and the replacement text is constant, the following awk script should do what you need:
BEGIN { normal = 1 }
/\/\/ ----------/ {
if (normal) {
normal = 0;
print "// **********";
print "// some replacement";
print "// text";
print "// that could have any";
print "// format";
print "// **********";
} else {
normal = 1;
next;
}
}
{
if (normal) print;
}
This prints everything it sees until it runs into the paragraph delimiter. When it sees the first one, it prints out the replacement paragraph. Until it sees the 2nd paragraph delimiter, it will print nothing. When it sees the 2nd paragraph delimiter, it will start printing lines normally again with the next line.
While you can technically do this from the command line, you may run into tricky shell quoting issues, especially if the replacement text has any single quotes. It may be easier to put the script in a file. Just put #!/usr/bin/awk -f (or whatever path which awk returns) at the top.
EDIT
To match multiple lines in awk, you'll need to use getline. Perhaps something like this:
/\/\/ ----------/ {
lines[0] = "// header";
lines[1] = "// comment";
lines[2] = "// to be replaced";
lines[3] = "// ----------";
linesRead = $0 "\n";
for (i = 0; i < 4; i++) {
getline line;
linesRead = linesRead line;
if (line != lines[i]) {
print linesRead; # print partial matches
next;
}
}
# print the replacement paragraph here
next;
}