Search Results

Yes, just put brackets:-

sed "s/^\(.........................\).*$/\1[...]/"

Try this to replace characters beyond 25th position:-

sed "s/^\(.........................\).*$/\1.../"

curl -s --connect-timeout 30 ftp://ftp2.bom.gov.au/anon/gen/fwo/IDA00003.html |
awk -F'<TD valign=top>| &nbsp;</TD>' '$0 ~ where {
printf "Today:\n%s\nTomorrow:\n%s\n", \
(length($8) >...

Isn't this what you want...

echo $peak $days_remaining |awk '{ printf "Per day: %.2f GB\n", (50-($1/1000000000))/$2 }'

Just added /$2 to your code

set `expr 50 - $peak / 1000000000` | echo $1 $days_remaining | awk '{ printf "Use %.2f GB per day of total %.2f GB free.\n", ($1 / $2), $1}'

set `expr 50 - $peak / 1000000000` | awk '{ printf "Use %.2f GB per day of total %.2f GB free.\n", ($1 / $days_remaining), $1}'