Extracting string between brackets

Question

How can I extract exact string between brackets?

What I tried is:

echo "test [test1] test" | grep -Po "(?=\[).*?(?=\])"

But the output is:

[test1

It should be:

test1

Better to use grep.

Answer 1

Use a lookbehind:

echo "test [test1] test" | grep -Po "(?<=\[).*?(?=\])"

Answer 2

awk should do too:

echo "test [test1] test" | awk -F"[][]" '{print $2}'
test1

Or sed

echo "test [test1] test" | sed 's/[^[]*\[\|\].*//g'
test1