How do I pipe output to date -d "value"?

Question

I have a date like 2014-01-30 05:04:27 GMT, and if I run date -d "2014-01-30 05:04:27 GMT", the output is in my server's timezone (Thu Jan 30 16:04:27 EST 2014).

With the use of grep and cut, I have extracted the date in GMT from a file. However, I am struggling to then convert this into my local time.

For example:

grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'['

Output: 2014-01-30 05:04:27 GMT

What can I add on the end, to pass that output to date -d?

Attempted:

grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[' | date -d

grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[' | date

grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[' | date -d "$1"

Answer 1

gmt="$(grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[')"
date -d "$gmt"

Or, if you prefer the pipeline format:

grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[' | { read gmt ; date -d "$gmt" ; }

The problem is that date does not use stdin. Thus, we have to capture the stdin into a variable (called gmt here) and then supply that on the command line to date.

Sample output from the second approach:

$ echo  "2014-01-30 05:04:27 GMT" | { read gmt ; date -d "$gmt" ; }
Wed Jan 29 21:04:27 PST 2014

Answer 2

If you're using GNU date from a sufficiently recent coreutils, there's date -f, from the help screen:

-f, --file=DATEFILE       like --date once for each line of DATEFILE

So your attempt 4 could have been:

$ grep "something" logfile.txt | grep "Succeeded" | cut -f1 -d'[' | date -f -

the last - stands for stdin.

Answer 3

Just use stdin as file:

command | date --file=-