mjhjmtu

inputs:

line1 with the PATTERN that contains ( ) 
line2 with the PATTERN that contains ( ) 
lineN with the PATTERN that contains ( ) 

outputs:

line1 with the PATTERN that contains ( ) ;
line2 with the PATTERN that contains ( ) ;
...
lineN with the PATTERN that contains ( ) ;

I tried this:

find . -name "test.txt" -print | xargs sed -i "/PATTERN/ s/$)/); /g"

but it didn't work.

perl -ipe 's/$/;/ if /PATTERN/'

This will add a ; at the end if the line contains PATTERN.

The $ matches the end of the line, so your pattern should be )$ instead of $) like in your example.

Also, you don't need xargs here, it's safer to use the -exec flag of fine:

find . -name test.txt -exec sed -i '/PATTERN/ s/)$/); /' '' +

If your version of find doesn't work with + at the end, then use ; instead (thank you @glenn-jackman):

find . -name test.txt -exec sed -i '/PATTERN/ s/)$/); /' '' ;

Finally, there's no need for the g flag in a s/something$// idiom, as there is only one occurrence of $ per line.

Assuming that PATTERN is actually ( ) and that something might go in between the ( ) and that they are not necessarily at the end of the line:

sed -i '/(.*)/ s/$/ ;/' test.txt

Using ex (which is equivalent to vi -e/vim -e).

One file:

ex +"g/local/s/$/;/g" -cwq foo.txt

All test.txt files recursively:

ex +"bufdo g/local/s/$/;/g" -cxa **/test.txt

^{Note: Make sure that globbing option (**) is enabled by: shopt -s globstar if your shell supports it.}

^{Note: The :bufdo command is not POSIX.}

Try:

sed --in-place '/PATTERN/s/.*/&;/' /path/to/file.txt

como seria en las lineas que no coicidan con la palabra