How to pipe only 1 $i?

motherboard · 03-16-2020, 10:01 AM

for i in mirror.foo ; do echo $i $i | rev ; done
oof.rorrim oof.rorrim

How do I get rev to pipe only the second $i?

The outcome should look like this

mirror.foo oof.rorrim

TenTenths · 03-16-2020, 10:13 AM

Code:

man echo

hint, -n is your friend.

TenTenths · 03-16-2020, 10:16 AM

Another method in bash is to make use of the $() construct to execute a command "in-line"

michaelk · 03-16-2020, 10:16 AM

There are multiple ways.
Command substitution is another hint.

Replied to late...

teckk · 03-16-2020, 03:55 PM

Quote:

for i in mirror.foo ; do echo $i $i

That will echo the same line twice. Why are you doing that?

Quote:

The outcome should look like this

mirror.foo oof.rorrim

Code:

mirror_foo="oof.rorrim"

for i in mirror_foo; do
    echo ""$i" "${!i}""
done

mirror_foo oof.rorrim

Is mirror.foo a variable or a file name?
. is not allowed in a bash variable.