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Date format conversation Linux

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# 1  
Old 10-24-2017
Date format conversation Linux
Hello,

i have hexadecimal number which need to be convert into datetime format.

manually i am able to using below ways.

converting hexdecimal to decimal

Code:
var1=`awk -F ',' '{print $4}' a.txt`
x=printf "%d\n" 0x$var1
z=`date -d @$x`

Can any one help me to make it in a single line execution command
where i can read each line of the file, pick 4th column from it(hexadecimal value) and convert it into decimal then to time.


Regards,
Last edited by Scott; 10-24-2017 at 06:31 PM.. Reason: Code tags
# 2  
Old 10-24-2017
Code:
while IFS=, read a b c d e ; do printf "%s\n" "$(date -d @$(printf "%d" 0x$d))" ; done < a.txt

This User Gave Thanks to rdrtx1 For This Post:
Chubler_XL
# 3  
Old 10-25-2017
Good approach! But - why the second printf?
Code:
while IFS=, read _ _ _ d _ ; do date -d @$(printf "%d" 0x$d); done < a.txt

would do as well, no?

And, if the file has one single line only, try
Code:
date -d @$(printf "%d" 0x$(cut -d, -f4 a.txt))

Last edited by RudiC; 10-27-2017 at 05:18 AM..
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