sed extracting group of digits

Question

I have tried to extract a number as given below but nothing is printed on screen:

echo \"This is an example: 65 apples\" | sed -n  \'s/.*\\([0-9]*\\) apples/\\1/         


        

Answer 1

echo "This is an example: 65 apples" | ssed -nR -e 's/.*?\b([0-9]*) apples/\1/p'

You will however need super-sed for this to work. The -R allows perl regexp.

Answer 2

$ echo "This is an example: 65 apples" | sed -r  's/^[^0-9]*([0-9]+).*/\1/'
65

Answer 3

It's because your first .* is greedy, and your [0-9]* allows 0 or more digits. Hence the .* gobbles up as much as it can (including the digits) and the [0-9]* matches nothing.

You can do:

echo "This is an example: 65 apples" | sed -n  's/.*\b\([0-9]\+\) apples/\1/p'

where I forced the [0-9] to match at least one digit, and also added a word boundary before the digits so the whole number is matched.

However, it's easier to use grep, where you match just the number:

echo "This is an example: 65 apples" | grep -P -o '[0-9]+(?= +apples)'

The -P means "perl regex" (so I don't have to worry about escaping the '+').

The -o means "only print the matches".

The (?= +apples) means match the digits followed by the word apples.

Answer 4

What you are seeing is the greedy behavior of regex. In your first example, .* gobbles up all the digits. Something like this does it:

echo "This is an example: 65144 apples" | sed -n  's/[^0-9]*\([0-9]\+\) apples/\1/p'
65144

This way, you can't match any digits in the first bit. Some regex dialects have a way to ask for non-greedy matching, but I don't believe sed has one.

Answer 5

A simple way for extracting all numbers from a string

echo "1213 test 456 test 789" | grep -P -o "\d+"

And the result:

1213
456
789