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Insert few lines above a match using sed, and within a perl file.

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# 1  
Insert few lines above a match using sed, and within a perl file.

Greetings all,

I am trying to match a string, and after that insert a few lines above that match.
The string is "Version 1.0.0". I need to insert a few lines ONLY above the first match (there are many Version numbers in the file). The rest of the matches must be ignored. The lines I need to insert are (above Version 1.0.0):
Version 2.0.0 
2012-01-11 - user_name
prod 2.0.0

This needs to be done in all files named VERSIONINFO recursively from the top level folder.
I have tried to do this in a perl file:
@readme_files = `find . -name "VERSIONINO"`;
foreach $readmes (@readme_files)        {
chomp ($readmes);
$ver_match = `sed -n '/^Version [0-9].[0-9].[0-9]/{p;q;}' "$readmes"`;
        if ($ver_match)
                {push (@vermatch,  $ver_match);
`sed  -i '/"$ver_match"/ i \Version 2.0.0\\n-------------\\n2012-01-15 - user_name\\nprod 2.0.0\\n' "$readmes"`;
        else {
                push (@no_vermatch, $readmes);
print "Versions found @vermatch\n";
print "Files with no version numbers @no_vermatch\n";

But the problem is with SED that does the actual insert. the error is:
sed: -e expression #1, char 15: unterminated address regex

sed, that picks out the first match works fine, the insert part does not. I tried escaping characters, but have not been successful. I believe I need to escape some characters in the sed insert command, but have been lost now.

Would appreciate any input to solve this.
Thanks in advance,

Last edited by Franklin52; 01-31-2012 at 03:22 AM.. Reason: Please use code tags for code and data samples, thank you
# 2  
Output in green.
$ cat inputfile
Version 1.0.0

Version 1.0.0

Version 1.0.0
$ sed '0,/Version 1.0.0/i Version 2.0.0\n-------------\n2012-01-11 - user_name\nprod 2.0.0' inputfile
Version 2.0.0
2012-01-11 - user_name
prod 2.0.0
Version 1.0.0

Version 1.0.0 

Version 1.0.0 

Last edited by balajesuri; 01-31-2012 at 02:56 AM..
# 3  
Please use code tags when posting sample input/output or code.

Your problem is the quotes:
sed  -i '/"$ver_match"/ i YourString'

The variable does not get expanded, because it's inside the single quotes, taken as literal. Turn the single quotes off and on:
sed  -i '/'"$ver_match"'/ i YourString'

and you should be good to go.

However this can be simplified quite a bit. All this perl stuff is not needed, since it only calls external tools (find, sed). You can do without:
v="Version 1.0"; 
str="Version 2.0.0\x0a-------------\x0a2012-01-15 - user_name\x0aprod 2.0.0"
find .  -name "VERSIONINFO" | xargs sed -i '/^'"$v"'/ i '"$str"

The \x0a is a hex code of the newline char.
xargs will feed all the files to sed as arguments, so sed is actually called like:
sed -i <command> ./path1/VERSIONINFO ./path2/VERSIONINFO ./path3/VERSIONINFO

Last edited by mirni; 01-31-2012 at 03:33 AM..

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