Grep: show only first match per line?

Hi,

can I make grep stop after finding the first occurrence of a regex in a line?

Given:

file with various regex patterns
file to be grep'ed

Sadly some of the regex patterns cannot be limited any further, so

grep -Eiof patterns.txt file.txt

(GNU grep 2.20) will give me possibly n hits for one line of the "source". But I only need the first hit per line.

Can I do this with grep?

And if grep is not my friend (in this case), how can I reach the goal instead? With an awk or sed loop?

Thank you!

Stephan

Show the input you have, and show the output you want.

grep is going to show the whole line unless the c or s option is used. You could used sed to insert escape sequences around your searched for term to make the first one stand out (apply a color or invert the display).

How about

awk '
NR == FNR       {SRCH = SRCH DL $0
                 DL = "|"
                 next
                }
match ($0, SRCH){print substr ($0, RSTART, RLENGTH)
                }
' patterns.txt file.txt

Thank you RudiC! That's exactly what I needed!
Many regards
Stephan