在工作中重新計劃

Question

有沒有辦法更改使用unix at命令發出的工作的日期？

我需要這樣做，因爲我的應用程序在同一時間安排了太多的作業，這會使機器停下來。

Answer 1

這可能是實現相關的，但是我的系統上我可以重命名，像這樣的工作：

$ # Bash 
$ sudo ls -l /var/spool/cron/atjobs 
-rwx------ 1 username daemon 3782 Nov 29 11:24 a00078014854e8 
$ atq 
120  Mon Nov 29 11:44:00 2010 a username 
$ printf "%x\n" $((16#14854e8 + 60*2)) # 2 hour delay 
1485560 
$ sudo mv /var/spool/cron/atjobs/a00078014854e8 /var/spool/cron/atjobs/a0007801485560 
$ atq 
120  Mon Nov 29 13:44:00 2010 a username 

在文件名中的最後8個十六進制數字是確定的時間從Unix紀元的分鐘數運行這項工作。按延遲的分鐘數遞增。

編輯：

下面是一個bash腳本來自動執行上述步驟。下面是一些例子運行：

創建工作：

$ date 
Mon Nov 29 20:00:00 CST 2010 
$ echo true | at now + 1 hour 
$ atq 
140  Mon Nov 29 21:00:00 2010 a username 

重新安排工作一小時後：

$ sudo atrs 140 60 
Job 140 in Queue "a" rescheduled 
from Mon Nov 29 21:00:00 CST 2010 
to Mon Nov 29 22:00:00 CST 2010 

重新計劃15分鐘前：

$ sudo atrs 140 -15 
Job 140 in Queue "a" rescheduled 
from Mon Nov 29 22:00:00 CST 2010 
to Mon Nov 29 21:45:00 CST 2010 

現在添加一天：

$ sudo atrs 140 $((60 * 24)) 
Job 140 in Queue "a" rescheduled 
from Mon Nov 29 21:45:00 CST 2010 
to Mon Nov 30 21:45:00 CST 2010 

您可以指定一個隊列：

$ sudo atrs -q b 141 120 

做一個預演：

$ sudo atrs -n 140 30 
Job 140 in Queue "a" 
Current job time: Mon Nov 30 21:45:00 2010 
Proposed job time: Mon Nov 30 22:15:00 2010 

這裏的腳本：

#!/bin/bash 
# atrs - reschedule at jobs 
# atrs [-n] [-q queue] job [-|+]minutes 

# by Dennis Williamson 2010-11-29 
# in response to http://stackoverflow.com/questions/4304631/rescheduling-an-at-job 

# for Bash 3.2 or greater 

# this script assumes that the last eight characters of the at job filename is 
# a sequence of hex digits representing the number of minutes starting at 
# the Unix epoch that is the time that the job is scheduled to be run 

LC_COLLATE=C 
export LC_TIME=C 
shopt -s nullglob 

mvcmd=/bin/mv 
datecmd=/bin/date 
GREP_OPTIONS= 
grepcmd=/bin/grep 
atqcmd=/usr/bin/atq 

atjobs=/var/spool/cron/atjobs 

declare -r tab=$'\t' 
declare -r NOEXIT=0 
declare -r EXIT=1 
# it's not necessary to bitmap the errors, but I just wanted to 
declare -r ERROPTS=1 
declare -r ERROARG=2 
declare -r ERRARGS=4 
declare -r ERRQUNM=8 
declare -r ERRNOJB=16 
declare -r ERRMVFL=32 
declare -r ERRNOCH=64 
declare -r ERRNINT=128 
declare -r DRYRUN=255 # not otherwise possible to reach this number 

queue=a 
err=0 

error() { 
    ((err |= ${2:-0})) 
    msg+="${3:+$3\n}" 
    if (($1 == $EXIT)) 
    then 
     printf "$msg" 
     printf "Usage: ${0##*/} [-n] [-q queue] job [-|+]minutes\n" 
     printf "  the default queue is a\n" 
     printf "  -n = dry run (default if not superuser)\n" 
     exit $err 
    else 
     return 

    fi 
} 

# Process options and arguments 
options=':q:nh' 
while getopts $options option 
do 
    case $option in 
     q ) queue=$OPTARG;; 
     n ) execute=1; ret=$DRYRUN;; # do dry run 
     h ) error $EXIT $DRYRUN;; 
     \?) if (((err & ERROPTS) != ERROPTS)) 
       then 
        error $NOEXIT $ERROPTS "Unknown option." 
       fi;; 
     * ) error $NOEXIT $ERROARG "Missing option argument.";; 
    esac 
done 

shift $(($OPTIND - 1)) 

if [[ ! $queue =~ ^[a-zA-Z=]$ ]] 
then 
    error $NOEXIT $ERRQUNM "Invalid queue name." 
fi 

if (($# != 2)) 
then 
    error $NOEXIT $ERRARGS "Job number and offset in minutes are required." 
fi 

if [[ $1 =~ ^[0-9]+$ ]] 
then 
    job=$1 
else 
    error $NOEXIT $ERRNINT "Job number must be a positive integer." 
fi 

if [[ $2 =~ ^[-+]?[0-9]+$ ]] 
then 
    minutes=$2 
else 
    error $NOEXIT $ERRNINT "Minutes must be an integer." 
fi 

if ((err != 0)) 
then 
    error $EXIT 
fi 

# make preparations 
if (($EUID == 0)) 
then 
    printf -v old "%05x" "$job" 
    prefix="$atjobs/$queue$old" 
    file=($prefix*) 
    if [[ -z $file || ! -e $file ]] 
    then 
     error $EXIT $ERRNOJB "Job not found." 
    fi 
    oldhex="${file#$prefix}" 
    oldminutes=$((16#$oldhex)) 
    newminutes=$((oldminutes + minutes)) 
    printf -v newhex "%08x" "$newminutes" 

    from=$($datecmd -d @"$(($oldminutes * 60))") 
    to=$($datecmd -d @"$((newminutes * 60))") 
else 
    if ((execute == 0)) 
    then 
     printf "You must be superuser to reschedule jobs. The job will be listed instead.\n" 
     execute=1 # do dry run 
     ret=$DRYRUN 
    fi 
fi 

# perform action 
if ((execute == 0)) 
then 
    if [[ $file != $prefix$newhex ]] 
    then 
     if $mvcmd "$file" "$prefix$newhex" 
     then 
      printf 'Job %s in Queue "%s" rescheduled\n' "$job" "$queue" 
      printf "from %s\n" "$from" 
      printf "to %s\n" "$to" 
     else 
      error $EXIT $ERRMVFL "Reschedule failed." 
     fi 
    else 
     error $EXIT $ERRNOCH "No change, times are the same." 
    fi 
else 
    jobdate=$($atqcmd -q $queue | $grepcmd "^$job$tab") 
    if [[ -n $jobdate ]] 
    then 
     jobdate=${jobdate#$job$tab} 
     jobdate=${jobdate%% $queue *} 
     newjobdate=$($datecmd +%c -d "$jobdate + $minutes minutes") 
     if [[ $jobdate != $newjobdate ]] 
     then 
      printf 'Job %s in Queue "%s"\n' "$job" "$queue" 
      printf "Current job time: %s\n" "$jobdate" 
      printf "Proposed job time: %s\n" "$newjobdate" 
     else 
      error $EXIT $ERRNOCH "Proposed time would result in no change." 
     fi 
    else 
     error $EXIT $ERRNOJB "Job not found." 
    fi 
fi 

exit $ret 

Answer 2

我不知道是否有一種方法來重新安排他們不同的時間，但你可以從at像這樣完全刪除它們：

首先列出工作：

$ at -l 
9944 2010-11-29 15:00 a dogbane 
9945 2010-11-29 15:00 a dogbane 

然後刪除它們：

$ at -d 9944 
$ at -d 9945 

然後，您可以重新創建它們，如果你想。

Answer 3

這將指派工作的一個新的ID它的新時間。

at -c OLDJOBID | at NEWTIME 
atrm OLDJOBID