0

I'm trying to do something like this in bash:

grep ( date | awk '{print "2006-" $6}' ) /some/file/here

But that syntax is incorrect.

The goal is to grep /some/file/here for the pattern 2006-2011 where 2011 is the current year.

3
grep "$(date | awk '{print "2006-" $6}')" /some/file/here

"..." holds its contents as one argument (even if there's whitespace).

$(...) is for "command substitution", where the stdout from running the embedded command will be put into place on the original command-line. (Another syntax, `...`, is also common but much harder to nest.)

Easier:

grep "$(date +'2006-%Y')" /some/file/here

Here, you use date's ability to format the output arbitrarily.

Note that none of these match any year in the range 2006-2011, they match the literal string "2006-2011". If you want to match any one year, let us know.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.