How to convert a hexadecimal to timestamp via bash script ?

nab_elf · 08-27-2013, 01:16 PM

Hello,
please, could you help me to resolve a mesterious problem when i convert time in hexadecimal format to humain format , i mean :

convert for example Ab153d to 00:00:00 with a bash script

Thank's in advance

duffmckagan · 08-27-2013, 01:37 PM

A quick google search of your requirement came up with the following result. Worth checking out I guess.

http://www.epochconverter.com/epoch/...-timestamp.php

smallpond · 08-27-2013, 02:14 PM

How's this?

Code:

perl -e '$foo=hex($ARGV[0]); @t=(localtime($foo))[2,1,0]; print $t[0],":",$t[1],":",$t[2],"\n"'  Ab153d
14:28:13

nab_elf · 08-28-2013, 05:25 AM

Hi,
thank's for replying ,

but is there any solution to do the same with bash script instead of perl ?

Thank's

nab_elf · 08-28-2013, 05:40 AM

Hello,
please find out in the attachment the resul when i lunch perl command. always the result = 0:0:0

druuna · 08-28-2013, 06:45 AM

I'm not entirely sure what your overall goal is, with that in mind.

Assuming that AB153D and 521DD201 are hex representations of linux epoch (seconds from January 1, 1970), have a look at this:

Code:

# hex input is 521dd201:
date --date @$( printf '%d\n' 0x521dd201) '+%Y-%m-%d %T'
2013-08-28 12:33:37

# hex input is Ab153d
date --date @$( printf '%d\n' 0xAb153d) '+%Y-%m-%d %T'
1970-05-10 19:28:13

The printf part converts from hex to decimal, which leaves you with epoch in seconds. This epoch can be given to the date command (--date @no_of_seconds), the last part ('+%Y-%m-%d %T') specifies the wanted output.

See the date manual page for details.

Firerat · 08-28-2013, 06:45 AM

Code:

printf "%d\n" 0x521DD201

date --date="@$(printf "%d\n" 0x521DD201)"

date "+%Y-%m-%d %H:%M:%S ( %s )" --date="@$(printf "%d\n" 0x521DD201)"

too slow

Firerat · 08-28-2013, 06:51 AM

and if you wanted to go the other way..

Code:

printf '0x%x\n' $(date +%s)