[SOLVED] print everything after $date

supulton · 02-21-2010, 05:46 AM

Say I have a line like such:

Code:

today      2010 Feb 21 test10, 12AM

How would I print everything AFTER "2010 Feb 21 "?

What I thought to do so far is to set a variable to the format used in the string like so:

Code:

date="$(date +%Y\ %b\ %e)"

So I know I have to use either awk, sed, or perhaps bash substrings to work with the variable $date, but as to how, I'm drawing a blank.

Any hints?

vikas027 · 02-21-2010, 05:50 AM

Quote:

Originally Posted by supulton

Say I have a line like such:

Code:

today      2010 Feb 21 test10, 12AM

How would I print everything AFTER "2010 Feb 21 "?

One suggestion is:

Code:

echo "today      2010 Feb 21 test10, 12AM" | awk '{print $5 "  " $6}'

ghostdog74 · 02-21-2010, 08:38 AM

Quote:

Originally Posted by vikas027

One suggestion is:

Code:

echo "today      2010 Feb 21 test10, 12AM" | awk '{print $5 "  " $6}'

what if there are more words after "2010 Feb 21" ? $5 and $6 would not be enough

ghostdog74 · 02-21-2010, 08:41 AM

Quote:

Originally Posted by supulton

Say I have a line like such:

Code:

today      2010 Feb 21 test10, 12AM

How would I print everything AFTER "2010 Feb 21 "?

What I thought to do so far is to set a variable to the format used in the string like so:

Code:

date="$(date +%Y\ %b\ %e)"

So I know I have to use either awk, sed, or perhaps bash substrings to work with the variable $date, but as to how, I'm drawing a blank.

Any hints?

Code:

$ line="today      2010 Feb 21 test10, 12AM"
$ echo ${line##*2010 Feb 21 }
test10, 12AM

Code:

$ var=$(date +%Y\ %b\ %e)
$ echo ${line##*$var }
test10, 12AM

supulton · 02-21-2010, 08:44 AM

Quote:

Originally Posted by ghostdog74

what if there are more words after "2010 Feb 21" ? $5 and $6 would not be enough

i realized this too and modified it:

Code:

echo "today      2010 Feb 21 two words, 12AM" | awk '{print substr($0, index($0,$5)) }'

this will print from the fifth word to the end of the line.

Code:

two words, 12AM

supulton · 02-21-2010, 08:46 AM

Quote:

Originally Posted by ghostdog74

Code:

$ line="today      2010 Feb 21 test10, 12AM"
$ echo ${line##*2010 Feb 21 }
test10, 12AM

Code:

$ var=$(date +%Y\ %b\ %e)
$ echo ${line##*$var }
test10, 12AM

this actually looks like a cleaner method. thanks!