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Can anyone help me to print UNIX epoch time to days,hours,min,sec ?

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Top Forums Shell Programming and Scripting Can anyone help me to print UNIX epoch time to days,hours,min,sec ?
# 8  
Old 10-05-2015
Maybe you mean something more like:
Code:
#!/bin/ksh
time="${1:-1441678454803}"
sec=$((time % 60))
time=$((time / 60))
min=$((time % 60))
time=$((time / 60))
hrs=$((time % 24))
days=$((time / 24))
[ $days -eq 1 ] && daysp="" || daysp=s
[ $hrs -eq 1 ] && hrsp="" || hrsp=s
[ $min -eq 1 ] && minp="" || minp=s
[ $sec -eq 1 ] && secp="" || secp=s
printf '%d day%s, %d hour%s, %d minute%s, %d second%s since the Epoch\n' \
    $days "$daysp" $hrs "$hrsp" $min "$minp" $sec "$secp"

which, for your sample value of 1441678454803 seconds since the Epoch produces the output:
Code:
16686093 days, 5 hours, 26 minutes, 43 seconds since the Epoch

and, if it is invoked with the operand 90061, produces the output:
Code:
1 day, 1 hour, 1 minute, 1 second since the Epoch

This User Gave Thanks to Don Cragun For This Post:
4walters
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