Bash脚本:两分之间的差异 [英] Bash script: difference in minutes between two times
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问题描述
标准日期
库是否可以执行此操作?
示例:
#!/ bin / bash
MPHR = 60分钟每小时。
CURRENT = $(date -u -d'2007-09-01 17:30:24''+%F%T.%N%Z')
TARGET = $( date -u -d'2007-12-25 12:30:00''+%F%T.%N%Z')
MINUTES = $(($(diff)/ $ MPHR ))
是否有一个更简单的方式这样做在hh中的时间和分钟:mm
解决方案
一个纯粹的 bash 解决方案:
old = 09:11
new = 17:22
#通过使用读取和拆分IFS
IFS =:read old_hour old_min<<< $ old
IFS =:读取小时最小值<<< $ new
#转换小时到分钟
#10#是避免错误的前导零
#通过告诉bash我们使用基础10
total_old_minutes = $((10#$ old_hour * 60 + 10#$ old_min))
total_minutes = $((10#$ hour * 60 + 10#$ min))
echo差别是$((total_minutes - total_old_minutes))分钟
另一个使用日期
(我们使用小时/分钟,所以日期不重要)
old = 09:11
new = 17:22
IFS =:read old_hour old_min<<<< $ old
IFS =:读取小时最小值<<< $ new
#从Unix EPOCH时间转换日期1970-01-01小时:分:00秒
sec_old = $(日期-d1970-01- 01 $ old_hour:$ old_min:00+%s)
sec_new = $(date -d1970-01-01 $ hour:$ min:00+%s)
差异是$(((sec_new - sec_old)/ 60))分钟
a href =http://en.wikipedia.org/wiki/Unix_time =noreferrer> http://en.wikipedia.org/wiki/Unix_time
I have two time strings; eg. "09:11" and "17:22" on the same day (format is hh:mm). How do I calculate the time difference in minutes between these two?
Can the standard date
library do this?
Example:
#!/bin/bash
MPHR=60 # Minutes per hour.
CURRENT=$(date -u -d '2007-09-01 17:30:24' '+%F %T.%N %Z')
TARGET=$(date -u -d'2007-12-25 12:30:00' '+%F %T.%N %Z')
MINUTES=$(( $(diff) / $MPHR ))
Is there a simpler way of doing this given the hour and minute in hh:mm
解决方案
A pure bash solution :
old=09:11
new=17:22
# feeding variables by using read and splitting with IFS
IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"
# convert hours to minutes
# the 10# is there to avoid errors with leading zeros
# by telling bash that we use base 10
total_old_minutes=$((10#$old_hour*60 + 10#$old_min))
total_minutes=$((10#$hour*60 + 10#$min))
echo "the difference is $((total_minutes - total_old_minutes)) minutes"
Another solution using date
(we work with hour/minutes, so the date is not important)
old=09:11
new=17:22
IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"
# convert the date "1970-01-01 hour:min:00" in seconds from Unix EPOCH time
sec_old=$(date -d "1970-01-01 $old_hour:$old_min:00" +%s)
sec_new=$(date -d "1970-01-01 $hour:$min:00" +%s)
echo "the difference is $(( (sec_new - sec_old) / 60)) minutes"
See http://en.wikipedia.org/wiki/Unix_time
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