how to use sed, awk, or gawk to print only what is matched?

Question

I see lots of examples and man pages on how to do things like search-and-replace using sed, awk, or gawk.

But in my case, I have a regular expression that I want to run

Answer 1

You can use awk with match() to access the captured group:

$ awk 'match($0, /abc([0-9]+)xyz/, matches) {print matches[1]}' file
12345

This tries to match the pattern abc[0-9]+xyz. If it does so, it stores its slices in the array matches, whose first item is the block [0-9]+. Since match() returns the character position, or index, of where that substring begins (1, if it starts at the beginning of string), it triggers the print action.

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With grep you can use a look-behind and look-ahead:

$ grep -oP '(?<=abc)[0-9]+(?=xyz)' file
12345

$ grep -oP 'abc\K[0-9]+(?=xyz)' file
12345

This checks the pattern [0-9]+ when it occurs within abc and xyz and just prints the digits.