在Shell脚本中将Cron Job设置为每月的第一个工作日 [英] Set Cron Job for 1st working day of every month in Shell Scripting

问题描述

我是脚本语言的新手，有人可以解释一下如何在第一工作日设置cron作业吗?

I'm new to scripting language, can anyone please explain how to set the cron job for 1st working day?

推荐答案

首先，运行date命令:

First, run the date command:

$ date '+%x'
> 12/29/2014

％x告诉日期以当前语言环境的格式显示今天的日期.使用完全相同的格式，将假日列表放入名为假日的文件中.例如:

%x tells date to display today's date in the format of the current locale. Using that exact same format, put a list of holidays in a file called holidays. For example:

$ cat holidays
> 01/01/2015
> 07/04/2015

接下来，创建以下shell脚本:

Next, create the following shell script:

#!/bin/sh
dom=$(date '+%d') # 01-31, day of month
year=$(date '+%Y') # four-digit year
month=$(date '+%m') # two-digit month

nworkdays=0
for d in $(seq 1 $dom)
do
    today=$(date -d "$year-$month-$d" '+%x') # locale's date representation (e.g. 12/31/99)
    dow=$(date -d "$year-$month-$d" '+%u')   # day of week: 1-7 with 1=Monday, 7=Sunday
    if [ "$dow" -le 5 ]  && grep -vq "$today" /path/holidays
    then
        workday=Yes
        nworkdays=$((nworkdays+1))
    else
        workday=
    fi
done
[ "$workday" ] && [ "$nworkdays" -eq 1 ] && /path/command

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