regexp (sed) suppress "no match" output

Question

I'm stuck on that and can't wrap my head around it: How can I tell sed to return the value found, and otherwise shut up?

It's really beyond me: Why would sed return the whole string if he found nothing? Do I have to run another test on the returned string to verify it? I tried using "-n" from the (very short) man page but it effectively suppresses all output, including matched strings.

This is what I have now :

echo plop-02-plop | sed -e 's/^.*\(.\)\([0-9][0-9]\)\1.*$/\2/' 

which returns 02 (and that is fine and dandy, thank you very much), but:

echo plop-02plop | sed -e 's/^.*\(.\)\([0-9][0-9]\)\1.*$/\2/' 

returns plop-02plop (when it should return this = "" nothing! Dang, you found nothing so be quiet! For crying out loud !!)

I tried checking for a return value, but this failed too ! Gasp !!

$ echo plop-02-plop | sed -e 's/^.*\(.\)\([0-9][0-9]\)\1.*$/\2/' ; echo $? 02 0 $ echo plop-02plop | sed -e 's/^.*\(.\)\([0-9][0-9]\)\1.*$/\2/' ; echo $? plop-02plop 0 $ 

This last one I cannot even believe. Is sed really the tool I should be using? I want to extract a needle from a haystack, and I want a needle or nothing..?

Answer 1

sed by default prints all lines.

What you want to do is

/patt/!d;s//repl/ 

IOW delete lines not matching your pattern, and if they match, extract particular element from it, giving capturing group number for instance. In your case it will be:

sed -e '/^.*\(.\)\([0-9][0-9]\)\1.*$/!d;s//\2/' 

You can also use -n option to suppress echoing all lines. Then line is printed only when you explicitly state it. In practice scripts using -n are usually longer and more cumbersome to maintain. Here it will be:

sed -ne 's/^.*\(.\)\([0-9][0-9]\)\1.*$/\2/p' 

There is also grep, but your example shows, why sed is sometimes better.

Answer 2

Perhaps you can use egrep -o?

input.txt:

blooody aaaa bbbb odor qqqq 

E.g.

sehe@meerkat:/tmp$ egrep -o o+ input.txt  ooo o o sehe@meerkat:/tmp$ egrep -no o+ input.txt  1:ooo 4:o 4:o 

Of course egrep will have slightly different (better?) regex syntax for advanced constructs (back-references, non-greedy operators). I'll let you do the translation, if you like the approach.