sed - how to remove everything but a defined pattern?

Question

I have to remove everything but 1, 2, or 3 digits (0-9, or 10-99, or 100) preceding % (I don't want to see the %, though) from another command's output and pipe it forward to another command. I know that

sed -n '/%/p'

will show only the line(s) containing %, but that's not what I want. How can I get rid of the rest of the unwanted text and leave only these numbers to then pipe them to another command?

Answer 1

If you're not completely tied to sed, this is exactly what grep -o does:

grep -o '[0-9]\{1,3\}%'

Answer 2

EDIT: I have misunderstood the OP and posted an invalid answer. I changed it to an answer that, I believe, would solve the problem in the more general scenario.

For a file such as the one below:

$ cat input
abc
123%
123
abc%
this is 456% and nothing more
456

Use sed -n -E 's/(^|.*[^0-9])([0-9]{1,3})%.*/\2/p' input

$  sed  -n -E 's/(^|.*[^0-9])([0-9]{1,3})%.*/\2/p' input
123
456

The -n flag makes sed to suppress automatic output of the lines. Then, we use the -E flag which will allow us to use extended regular expressions. (In GNU sed, the flag is not -E but instead is -r).

Now comes the s/// command. The group (^|.*[^0-9]) matchs either a beginning of line (^) or a series of zero or more chars (.*) ending in a non-digit char ([^0-9]). [0-9]\{1,3\} just matches one to three digits and is bound to a group (by the ( and ) group delimiters) if the group is preceded by (^|.*[^0-9]) and followed by %. Then .* matches everything before and after this pattern. After this, we replace everything by the second group (([0-9]{1,3})) using the backreference \2. Since we passed -n to sed, nothing would be printed but we passed the p flag to the s/// command. The result is that if the replacement is executed then the resulted line is printed. Note the p is a flag of s///, not the p command, because it comes just after the last /.