Dotan writes: > Is there a way to grep the output of strings in order to only show > lines that contain words found in the aspell dictionary?
Try this: #!/bin/bash strings "$1" | while read line do if [ ` echo "$line" | sed -e 's/[^a-zA-Z ]//g' | wc -m` -lt 6 ] then continue fi echo "$line" | sed -e 's/[^a-zA-Z ]//g' | aspell -W1 --dont-guess --dont-suggest -a | grep -q '\*' if [ $? -eq 0 ] then echo "$line" fi done Lines with fewer than 6 letters are rejected. You might want to adjust that value, as well as the character class in the sed scripts. -- John Hasler -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of "unsubscribe". Trouble? Contact listmas...@lists.debian.org
