How do I covert this date format to epoch?

Question

I know the command is date +%s -d "xxx"

But how do I convert 11/03/2018 02:44:58 to epoch ?

root@AngelBeats:~# date +%s -d "11/03/2018 02:44:58"
date: invalid date '11/03/2018 02:44:58'

Yuvraj Jaiswal · Answer 1 · 2018-03-11T06:30:34.533

2

Edit: For Linux

date --date="11/03/2018 8:15:00" +"%s"

1520736300

This should do it.(For OSX Systems)

date -j -f "%d/%m/%Y %H:%M:%S" "11/03/2018 8:15:00" +"%s"

1520736300

The "-f" specifier can be used to set the format.

edited Mar 11 '18 at 06:30

answered Mar 10 '18 at 15:57

Yuvraj Jaiswal

1

Should be `date -j -f "%d/%m/%Y %H:%M:%S" "11/03/2018 8:15:00" +"%s"` to get the epoch. – codeforester Mar 10 '18 at 16:03
@codeforester my bad, didn't see. Thanks a lot. Edited – Yuvraj Jaiswal Mar 10 '18 at 16:05
`date: invalid option -- 'j' Try 'date --help' for more information.` I'm using manjaro – Joe Mar 10 '18 at 16:09
try this: ` date --date="11\03\2018 8:15:00" +"%s" – Yuvraj Jaiswal Mar 10 '18 at 16:18

score 1 · Answer 2 · answered Mar 10 '18 at 16:15

1

-j is for BSD systems. On Linux, use this:

date -d "11/03/2018 8:15:00" +"%s"

answered Mar 10 '18 at 16:15

codeforester

2 Answers2