LinuxQuestions.org - [SOLVED] print everything after $date

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supulton

02-21-2010 05:46 AM

print everything after $date

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 (Post 3871318)

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 (Post 3871321)

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 (Post 3871318)

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 (Post 3871409)

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 (Post 3871411)

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!

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