Using sed, how do you print the first 'N' characters of a line?

Question

With sed, what is a one liner to print the first n characters. I am doing the following.

grep -G 'defn -test.*' OctaneFullTest.clj | sed ....

Answer 1

Don't use sed, use cut.

grep .... | cut -c 1-N

If you MUST use sed:

grep ... | sed -e 's/^\(.\{12\}\).*/\1/'

Answer 2

For more complex processing, I recommend writing a quick awk script. It's a handy skill to have from time to time. Sorry no code right now. But that is easy in awk.

Answer 3

Strictly with sed:

grep ... | sed -e 's/^\(.\{N\}\).*$/\1/'

Answer 4

don't have to use grep either

an example:

sed -n '/searchwords/{s/^\(.\{12\}\).*/\1/g;p}' file

Answer 5

It works just wonderfull thank you !!!!!