我见过一些解决方案，包括监视和简单地在后台运行循环（和睡眠）脚本，但没有什么是理想的。
我有一个需要每 15 秒运行一次的脚本，而且由于 cron 不支持秒数，所以我只能找出其他的东西。
在 unix 上每 15 秒运行一次脚本的最强大和最有效的方法是什么？该脚本还需要在重新启动后运行。
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148750 次
9 回答
295
如果您坚持从 cron 运行脚本：
* * * * * /foo/bar/your_script
* * * * * sleep 15; /foo/bar/your_script
* * * * * sleep 30; /foo/bar/your_script
* * * * * sleep 45; /foo/bar/your_script
并将您的脚本名称和路径替换为 /foo/bar/your_script
于 2009-06-23T18:28:57.697 回答
77
我会使用 cron 每分钟运行一个脚本，并让该脚本运行你的脚本四次，运行之间有 15 秒的睡眠。
（假设您的脚本运行速度很快 - 如果没有，您可以调整睡眠时间。）
这样，您就可以享受cron15 秒的运行时间和所有好处。
编辑：另请参阅下面的@bmb 评论。
于 2009-06-23T18:19:47.087 回答
15
上面的修改版本：
mkdir /etc/cron.15sec
mkdir /etc/cron.minute
mkdir /etc/cron.5minute
添加到 /etc/crontab：
* * * * * root run-parts /etc/cron.15sec > /dev/null 2> /dev/null
* * * * * root sleep 15; run-parts /etc/cron.15sec > /dev/null 2> /dev/null
* * * * * root sleep 30; run-parts /etc/cron.15sec > /dev/null 2> /dev/null
* * * * * root sleep 45; run-parts /etc/cron.15sec > /dev/null 2> /dev/null
* * * * * root run-parts /etc/cron.minute > /dev/null 2> /dev/null
*/5 * * * * root run-parts /etc/cron.5minute > /dev/null 2> /dev/null
于 2013-10-23T14:44:40.690 回答
14
不会在后台运行它吗？
#!/bin/sh
while [ 1 ]; do
echo "Hell yeah!" &
sleep 15
done
这几乎是最有效的。重要的部分仅每 15 秒执行一次，脚本在其余时间休眠（因此不会浪费周期）。
于 2009-06-23T18:28:41.283 回答
1
我写了一个比 cron 更快的调度程序。我还实施了重叠防护。如果前一个仍在运行，您可以将调度程序配置为不启动新进程。看看https://github.com/sioux1977/scheduler/wiki
于 2015-09-15T10:40:08.243 回答
0
使用nanosleep(2)。它使用timespec用于以纳秒精度指定时间间隔的结构。
struct timespec {
time_t tv_sec; /* seconds */
long tv_nsec; /* nanoseconds */
};
于 2013-01-15T11:39:04.270 回答
0
#! /bin/sh
# Run all programs in a directory in parallel
# Usage: run-parallel directory delay
# Copyright 2013 by Marc Perkel
# docs at http://wiki.junkemailfilter.com/index.php/How_to_run_a_Linux_script_every_few_seconds_under_cron"
# Free to use with attribution
if [ $# -eq 0 ]
then
echo
echo "run-parallel by Marc Perkel"
echo
echo "This program is used to run all programs in a directory in parallel"
echo "or to rerun them every X seconds for one minute."
echo "Think of this program as cron with seconds resolution."
echo
echo "Usage: run-parallel [directory] [delay]"
echo
echo "Examples:"
echo " run-parallel /etc/cron.20sec 20"
echo " run-parallel 20"
echo " # Runs all executable files in /etc/cron.20sec every 20 seconds or 3 times a minute."
echo
echo "If delay parameter is missing it runs everything once and exits."
echo "If only delay is passed then the directory /etc/cron.[delay]sec is assumed."
echo
echo 'if "cronsec" is passed then it runs all of these delays 2 3 4 5 6 10 12 15 20 30'
echo "resulting in 30 20 15 12 10 6 5 4 3 2 executions per minute."
echo
exit
fi
# If "cronsec" is passed as a parameter then run all the delays in parallel
if [ $1 = cronsec ]
then
$0 2 &
$0 3 &
$0 4 &
$0 5 &
$0 6 &
$0 10 &
$0 12 &
$0 15 &
$0 20 &
$0 30 &
exit
fi
# Set the directory to first prameter and delay to second parameter
dir=$1
delay=$2
# If only parameter is 2,3,4,5,6,10,12,15,20,30 then automatically calculate
# the standard directory name /etc/cron.[delay]sec
if [[ "$1" =~ ^(2|3|4|5|6|10|12|15|20|30)$ ]]
then
dir="/etc/cron.$1sec"
delay=$1
fi
# Exit if directory doesn't exist or has no files
if [ ! "$(ls -A $dir/)" ]
then
exit
fi
# Sleep if both $delay and $counter are set
if [ ! -z $delay ] && [ ! -z $counter ]
then
sleep $delay
fi
# Set counter to 0 if not set
if [ -z $counter ]
then
counter=0
fi
# Run all the programs in the directory in parallel
# Use of timeout ensures that the processes are killed if they run too long
for program in $dir/* ; do
if [ -x $program ]
then
if [ "0$delay" -gt 1 ]
then
timeout $delay $program &> /dev/null &
else
$program &> /dev/null &
fi
fi
done
# If delay not set then we're done
if [ -z $delay ]
then
exit
fi
# Add delay to counter
counter=$(( $counter + $delay ))
# If minute is not up - call self recursively
if [ $counter -lt 60 ]
then
. $0 $dir $delay &
fi
# Otherwise we're done
于 2014-01-09T14:36:08.170 回答
0
自从我之前的回答以来，我想出了另一个不同的解决方案，也许更好。此代码允许进程以微秒精度每分钟运行 60 次以上。您需要 usleep 程序来完成这项工作。应该好到每秒50次。
#! /bin/sh
# Microsecond Cron
# Usage: cron-ms start
# Copyright 2014 by Marc Perkel
# docs at http://wiki.junkemailfilter.com/index.php/How_to_run_a_Linux_script_every_few_seconds_under_cron"
# Free to use with attribution
basedir=/etc/cron-ms
if [ $# -eq 0 ]
then
echo
echo "cron-ms by Marc Perkel"
echo
echo "This program is used to run all programs in a directory in parallel every X times per minute."
echo "Think of this program as cron with microseconds resolution."
echo
echo "Usage: cron-ms start"
echo
echo "The scheduling is done by creating directories with the number of"
echo "executions per minute as part of the directory name."
echo
echo "Examples:"
echo " /etc/cron-ms/7 # Executes everything in that directory 7 times a minute"
echo " /etc/cron-ms/30 # Executes everything in that directory 30 times a minute"
echo " /etc/cron-ms/600 # Executes everything in that directory 10 times a second"
echo " /etc/cron-ms/2400 # Executes everything in that directory 40 times a second"
echo
exit
fi
# If "start" is passed as a parameter then run all the loops in parallel
# The number of the directory is the number of executions per minute
# Since cron isn't accurate we need to start at top of next minute
if [ $1 = start ]
then
for dir in $basedir/* ; do
$0 ${dir##*/} 60000000 &
done
exit
fi
# Loops per minute and the next interval are passed on the command line with each loop
loops=$1
next_interval=$2
# Sleeps until a specific part of a minute with microsecond resolution. 60000000 is full minute
usleep $(( $next_interval - 10#$(date +%S%N) / 1000 ))
# Run all the programs in the directory in parallel
for program in $basedir/$loops/* ; do
if [ -x $program ]
then
$program &> /dev/null &
fi
done
# Calculate next_interval
next_interval=$(($next_interval % 60000000 + (60000000 / $loops) ))
# If minute is not up - call self recursively
if [ $next_interval -lt $(( 60000000 / $loops * $loops)) ]
then
. $0 $loops $next_interval &
fi
# Otherwise we're done
于 2014-01-23T15:31:58.397 回答
-1
为避免可能的执行重叠，请使用该线程中描述的锁定机制。
于 2013-01-23T11:20:33.350 回答