bash 脚本在 5 分钟内运行
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bash script to run in 5 minutes
提问by James Nine
I just want a bash script to run 5 minutes after it's called. What am I doing wrong?
我只想让 bash 脚本在调用后运行 5 分钟。我究竟做错了什么？
I have the command:
我有命令：
/path/to/my/script | at now + 5 min
And yet the script runs right away every time.
然而，脚本每次都会立即运行。
回答by Hyman Kelly
You are executing the script immediately and sending its output into at. You need to send the name of the script itself into at:
您正在立即执行脚本并将其输出发送到at. 您需要将脚本本身的名称发送到at：
echo /path/to/my/script | at now + 5 min
回答by Amir Afghani
how about:
怎么样：
sleep 300 && /path/to/my/script
回答by oddenodin
at -f /path/to/my/script -t now +5 minutes
This should work as far as scheduling a script to run at a specific time. For any more information on the "at" command try linuxmanpages.com. I may be wrong thought ( currently not at a linux system to test ).
这应该可以安排脚本在特定时间运行。有关“ at”命令的更多信息，请访问linuxmanpages.com。我的想法可能是错误的（目前不是在 linux 系统上进行测试）。
Good luck anyways
还是祝你好运
回答by Femaref
Commands are evaluated left to right, so first your script gets executed, the output of it will be piped to the at command, this is normal behaviour. Have look at at the at man pagesfor more information.
命令从左到右计算，所以首先你的脚本被执行，它的输出将通过管道传输到 at 命令，这是正常行为。查看 at手册页了解更多信息。
回答by Andrew Cooper
The problem is you're running the script and piping the output to the atcommand. What you need to do is run the atcommand with the path to your script as a parameter. I'm not sure of the syntax, but at -hor man atshould help.
问题是您正在运行脚本并将输出通过管道传输到at命令。您需要做的是at使用脚本的路径作为参数运行命令。我不知道语法的，但at -h还是man at应该有所帮助。
回答by edmondscommerce
try this
尝试这个
sys.scheduled_run /path/to/my/script 5
sys.scheduled_run /path/to/my/script 5
main function
主功能
function sys.scheduled_run(){
local PATH_TO_ACTION MINS SLEEPTIME
PATH_TO_ACTION=
MINS=
SLEEPTIME=$(($MINS * 60))
echo "Sleeping for $MINS minute ($SLEEPTIME seconds) and then running $PATH_TO_ACTION"
ui.countdown $SLEEPTIME
$PATH_TO_ACTION
echo "Done"
if [ "REPEAT" == "" ]
then
echo "Going for Repeat"
sys.scheduled_run "$@"
fi
}
countdown function
倒计时功能
function ui.countdown(){ #USAGE ui.countdown 60 countdown for 60 seconds
local SECONDS=
local START=$(date +%s)
local END=$((START + SECONDS))
local CUR=$START
while [[ $CUR -lt $END ]]
do
CUR=$(date +%s)
LEFT=$((END-CUR))
printf "\r%02d:%02d:%02d" \
$((LEFT/3600)) $(( (LEFT/60)%60)) $((LEFT%60))
sleep 1
done
echo " "
}
