Ubuntu – how to use sed to replace from last 3 characters from string

command linesed

I have the issue to extract the x amount of characters from string and i have the difficulty enlarging the capability of the expression.

I have tried to use the sed command.

If i have the string example.com.db i need to keep "example.com",for that i use

sed 's/\.db//g'

How to change expression so that i can use these kind of files too example.db.db.com.db.

in this case the end result i need is example.db.db.com

you can use:

's/.db$//'

alternatively you can use:

's/...$//'

to remove the last 3 characters and

's/\.[^\.]*$//'

to remove the everything after the last dot

There is a -F option:

-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched.  (-F is specified by POSIX.)

This will disable searching by regular expressions and interpreting characters like * or .. Example:

$ cat test 
a**__
$ grep -F 'a**__' test 
a**__

The following command should work for what you want:

sed "s/this is test/$(cat bar)/" foo

If foo contain more then one line, then you can use:

sed "s/this is test/$(sed -e 's/[\&/]/\\&/g' -e 's/$/\\n/' bar | tr -d '\n')/" foo

or:

sed -e '/this is a test/{r bar' -e 'd}' foo

To make the change in foo file, use sed -i.