[SOLVED] Converting date feed into different date format

keif · 07-08-2014, 08:09 AM

Hello all,

I'm grabbing information from the expiration date of java certificates from a Linux box, and I set the columns to come back as follows:

Code:

Apr 09 2014
Mar 07 2014
Mar 06 2017
Apr 10 2015
Apr 07 2014

I have these in five different variables. I would like to feed each of them into a command that converts it into a different date format, if possible.

Here is the format I have in mind:

Code:

4-9-2014
3-7-2014
3-6-2017
4-10-2015
4-7-2014

The only real difference besides the alphabetic change is the elimination of the '0s' in the day column.

If anyone knows a way to do this it would be greatly appreciated.

Thanks!

TenTenths · 07-08-2014, 08:19 AM

What have you tried so far?

syg00 · 07-08-2014, 08:20 AM

"man date" - will do all the formatting you require, including eliminating the zero padding.

keif · 07-08-2014, 08:26 AM

Quote:

Originally Posted by TenTenths

What have you tried so far?

I found a way to do this:

Code:

echo "$variable" | awk '{print "date -d\""$1FS$2FS$3"\" +%m-%d-%Y"}' | bash

Thanks for the push!

keif · 07-08-2014, 08:33 AM

Quote:

Originally Posted by syg00

"man date" - will do all the formatting you require, including eliminating the zero padding.

Actually, that would be the following to eliminate the padding for the day:

Code:

echo "$variable" | awk '{print "date -d\""$1FS$2FS$3"\" +%m-%-d-%Y"}' | bash

Thanks for using the word "padding". That sent me right where I need to go :-)

pan64 · 07-08-2014, 08:39 AM

that will not remove 0s from the day coloumn.
also
date -d "$variable" '+%m-%-d-%Y'
will make the same thing.