bash 使用 date 命令比较时间
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Compare time using date command
提问by vehomzzz
Say I want a certain block of a bash script executed only if it is between 8 am (8:00) and 5 pm (17:00), and do nothing otherwise. The script is running continuously.
假设我希望某个 bash 脚本块仅在上午 8 点 (8:00) 和下午 5 点 (17:00) 之间执行，否则不执行任何操作。脚本一直在运行。
So far I am using the datecommand.
到目前为止，我正在使用该date命令。
How to use it to compare it to the time range?
如何使用它与时间范围进行比较？
回答by Cascabel
Just check if the current hour of the day is between 8 and 5 - since you're using round numbers, you don't even have to muck around with minutes:
只需检查一天中的当前小时是否在 8 到 5 之间 - 由于您使用的是整数，因此您甚至不必纠结于分钟：
hour=$(date +%H)
if [ "$hour" -lt 17 -a "$hour" -ge 8 ]; then
# do stuff
fi
Of course, this is true at 8:00:00 but false at 5:00:00; hopefully that's not a big deal.
当然，这在 8:00:00 为真，但在 5:00:00 为假；希望这不是什么大不了的事。
For more complex time ranges, an easier approach might be to convert back to unix time where you can compare more easily:
对于更复杂的时间范围，更简单的方法可能是转换回 unix 时间，以便您可以更轻松地进行比较：
begin=$(date --date="8:00" +%s)
end=$(date --date="17:00" +%s)
now=$(date +%s)
# if you want to use a given time instead of the current time, use
# $(date --date=$some_time +%s)
if [ "$begin" -le "$now" -a "$now" -le "$end" ]; then
# do stuff
fi
Of course, I have made the mistake of answering the question as asked. As seamus suggests, you could just use a cron job - it could start the service at 8 and take it down at 5, or just run it at the expected times between.
当然，我犯了错误地回答问题。正如 seamus 所建议的那样，您可以只使用 cron 作业 - 它可以在 8 点启动服务并在 5 点关闭它，或者只是在预期的时间间隔运行它。
回答by seamus
Why not just use a cron job?
为什么不直接使用 cron 作业？
Otherwise
除此以外
if [[ `date +%H` -ge 8 && `date +%H` -lt 17 ]];then
do_stuff()
fi
will do the job
会做这份工作
