UNIX for Dummies Questions & Answers

Hi.

Can I extract the last 3 characters from a given string using sed ?

Why the following doesn't work (it prints the full string) :

echo "abcd" | sed '/\.\.\.$/p' doesn't work ?

output: abcd


Thanks in advance,
435 Gavea.

Your command looks for 3 explicit dots at the end of a line, if they exist, the line is printed. But you did not use -n, so the line gets printed anyway.

You want this:
sed 's/.*\(...\)$/\1/'

Hehe, obviosly \. was intended to be a real dot character...
I was kinda desperate , so I tried this out...

But, let me see If I got it:

You're substituting any characteres till the last 3 ones - (...) means exactly what ? to look for the pattern till (excluding) the last 3 characters ?

And what means the \1 ?

Why didn't you subtituted this characters with nothing ?

Thanks in advance,
435 Gavea

No, .*...$ matches the entire line, provided that at least 3 characters exist. And the entire line is replaced with \1. \1 is the first saved field. A saved field is stuff between \( and \).

Ok, but isn't any way to return directly the 3 characters ? I mean, you can't select directly these 3 characters using the right pattern ?

Or there isn't other way unless you select the entire line first... ?




435 Gavea.

Um, my method does select the 3 rightmost characters and return them.

There are other sed solutions, but I think this approach is best if sed must be used.

Try it,

echo abcd | sed -n 's/\(^.[^$]*\)\(.\{3\}$\)/\2/p'