This should work for you:
awk '/^abcd/{p++;if(p==1) print}/^DDD/{q++;if(q==8||q==9)print}' file
質問
I have a text file and I want to print the 1st line starting with pattern "abcd" and also the 8th and 9th lines which are started with "DDD" pattern. Any sed or awk for doing that?
解決
This should work for you:
awk '/^abcd/{p++;if(p==1) print}/^DDD/{q++;if(q==8||q==9)print}' file
他のヒント
perl -lne 'push @{ $h->{$1} }, $_ if /.*(abcd|DDD).*/;END{print join "\n", @{ $h->{abcd}}[0]; print join "\n", @{ $h->{DDD}}[7..8];}' file
or decomposed :
perl -lne '
push @{ $h->{$1} }, $_ if /.*(abcd|DDD).*/;
END{
print join "\n", @{ $h->{abcd} }[0], @{ $h->{DDD} }[7..8];
}
' file