如何添加Linux格式的时间输出

假设时间戳是

HH:MM:SS[:MILLISEC]

你一定要按照Ed和我们的要求解释你的时间戳的规格.

With bash

#!/bin/bash

timestamps=( $(
    showjobs -u rsingh70 -n 365 |
        awk '/CPUTime/{print $NF}'
))

for timestamp in "${timestamps[@]}"; do
    IFS=: read -ra a <<< "$timestamp"
    if ((${#a[@]} == 4)); then
        ((seconds+=10#${a[0]}*3600+10#${a[1]}*60+10#${a[2]}*1+10#${a[3]}/1000))
    else
        ((seconds+=10#${a[0]}*3600+10#${a[1]}*60+10#${a[2]}))
    fi
done

days=$((seconds/86400))
hours=$((seconds/3600%24))
minutes=$((seconds/60%60))
seconds=$((seconds%60))
echo "Total time is $days days $hours hours $minutes minutes $seconds secondes"

With awk

重新使用/borrow 来自Glenn的END部分优雅代码

showjobs -u rsingh70 -n 365 | awk '
    /CPUTime/ {
        split($NF, a, ":")
        if (NF == 5) {
            seconds+=a[1]*3600+a[2]*60+a[3]*1+a[4]/1000
        }
        else {
            seconds+=a[1]*3600+a[2]*60+a[3]
        }
    }
    END {
        s = seconds % 60; mins = (seconds - s)/60
        m = mins % 60; hrs = (mins - m)/60
        h = hrs % 24
        days = (hrs - h)/24
        printf "%d days %02d hours %02d minutes %02d seconds\n", days, h, m, s
  }
'

Output

12 days 11 hours 42 minutes 34 seconds