Print all words that start with a %, one to a line.

[jfm1]

Hello,

I have a log where every line has a word in it that starts with a % sign. I need to print each line, but only the word that starts with %, no other words or characters.

Thanks for your help,

Joe

 

[marsd]

Hey, Joe

It's easier just to use sed for this one.

cat logfile | sed '/^\%/!d'

Will give you the output that you want.

 

[marsd]

Sorry, I misunderstood what you wanted.
Assuming the wortds are separated by a space
then you could write this in awk and it would
work.

cat logfile | awk ' {
split($0, per, " &quot
for (i in per)
if ($i ~ /^\%/) {
print $i
}
}'

Bye now.

 

[jfm1]

I tried that but in csh I got "d': Event not found" as an error msg. But I tried combining sed and awk and finally got what I wanted:

cat logfile | sed 's/.*\(%\)/\1/' | awk '{print $1}'

This listed the "%words" I was looking for. Thanks for the kick start.

Joe