如何让unix脚本每15秒运行一次？

时间：2009-06-23 18:17:59

标签： unix cron command sleep

我已经看过一些解决方案，包括监视和在后台运行循环（和休眠）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，我只能找出别的东西。

在unix上每15秒运行一次脚本最强大有效的方法是什么？脚本也需要在重启后运行。

9 个答案:

答案 0 :(得分：280)

如果你坚持从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

答案 1 :(得分：75)

我会每分钟使用cron运行一个脚本，并使该脚本运行您的脚本四次，并在运行之间休眠15秒。

（假设您的脚本可以快速运行 - 如果没有，您可以调整睡眠时间。）

这样，您可以获得cron以及15秒运行期间的所有好处。

修改：另请参阅下面的@ bmb评论。

答案 2 :(得分：14)

以上修改版：

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

答案 3 :(得分：13)

不会在后台运行吗？

#!/bin/sh
while [ 1 ]; do
    echo "Hell yeah!" &
    sleep 15
done

这和它一样高效。重要的部分每15秒执行一次，脚本在剩下的时间内休眠（因此不会浪费周期）。

答案 4 :(得分：1)

我写的调度程序比cron快。我也实施了一个重叠的守卫。您可以将调度程序配置为在前一个进程仍在运行时不启动新进程。看看https://github.com/sioux1977/scheduler/wiki

答案 5 :(得分：0)

使用nanosleep(2)。它使用结构timespec，用于指定具有纳秒精度的时间间隔。

struct timespec {
           time_t tv_sec;        /* seconds */
           long   tv_nsec;       /* nanoseconds */
       };

答案 6 :(得分：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

答案 7 :(得分：0)

自从我之前的回答以来，我提出了另一种不同的解决方案，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。你需要使用usleep程序来完成这项工作。应该是每秒高达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

答案 8 :(得分：-2)

为避免执行可能重叠，请使用thread。

中所述的锁定机制
相关问题
最新问题