bash 今天的日期，在 shell 脚本中减去 X 天
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Today's date, minus X days in shell script
提问by gcubed
I need to create three variables, each for Year, Month, and Day for Today's date, minus X number of days. For this question I'll choose a random amount of days: 222.
我需要创建三个变量，每个变量代表今天的年、月和日，减去 X 天数。对于这个问题，我将选择一个随机天数：222。
So if:
因此，如果：
TodayYear=`date +%Y`
TodayMonth=`date +%m`
TodayDay=`date +%d`
What I want is 222 days before this.
我想要的是在此之前的 222 天。
222days_before_TodayYear=???
222days_before_TodayMonth=???
222days_before_TodayDay=???
Edit: Need 222 workingdays instead 222 regular days.
编辑：需要 222 个工作日而不是 222 个正常工作日。
回答by sampson-chen
For GNU date:
对于 GNU date：
date_222days_before_TodayYear=$(date --date="222 days ago" +"%Y")
date_222days_before_TodayMonth=$(date --date="222 days ago" +"%m")
date_222days_before_TodayDay=$(date --date="222 days ago" +"%d")
For BSD date::
对于 BSD date::
If you are using OS X or FreeBSD, use the following instead because BSD date is different from GNU date:
如果您使用的是 OS X 或 FreeBSD，请改用以下内容，因为 BSD 日期与 GNU 日期不同：
date_222days_before_TodayYear=$(date -j -v-222d +"%Y")
date_222days_before_TodayMonth=$(date -j -v-222d +"%m")
date_222days_before_TodayDay=$(date -j -v-222d +"%d")
Source: BSD date manual page
来源：BSD 日期手册页
Note:
笔记：
In bashand many other languages, you cannot start a variable name with a numerical character, so I prefixed them with date_for you.
在bash许多其他语言中，您不能以数字字符开头变量名，因此我date_为您添加了前缀。
Second Update:New requirement - Using 222 Working days instead of 222 Regular days:
第二次更新：新要求 -使用 222 个工作日而不是 222 个正常工作日：
(Assumption:Not considering statutory holidays, because that just gets far beyond the scope of what I can help you with in a shell script:)
（假设：不考虑法定假日，因为这远远超出了我在 shell 脚本中可以帮助您的范围：）
Consider 222 working days:
考虑 222 个工作日：
- 5 working days per week, that is
floor(222/5) == 44 weeks 44 weeks * 7 days per week == 308 days- Extra days leftover:
222 % 5 == 2 - Therefore
222 working days == 310 regular days
- 每周 5 个工作日，即
floor(222/5) == 44 weeks 44 weeks * 7 days per week == 308 days- 剩余天数：
222 % 5 == 2 - 所以
222 working days == 310 regular days
But, there is a catch! If the number of regular days is 308or some multiple of 7, then we would have been fine, because any multiple of 7-days ago from a working day is still a working day. So we need to consider whether today is a Monday or a Tuesday:
但是有一个问题！如果常规天数是308或 的倍数7，那么我们就可以了，因为 7 天前的任何一个工作日的倍数仍然是一个工作日。所以我们需要考虑今天是星期一还是星期二：
- If today is a Monday, we'd get Saturday where we wanted Thursday
- If today is a Tuesday, we'd get Sunday where we wanted Friday
- 如果今天是星期一，我们会在星期六到达我们想要的星期四
- 如果今天是星期二，我们会在星期天得到我们想要的星期五
So you see we need an additional offset of 2 more days if today is either Monday or Tuesday; so let's find that out first before we proceed:
所以你看，如果今天是星期一或星期二，我们需要额外的 2 天的抵消；所以在我们继续之前，让我们先找出来：
#!/bin/bash
# Use 310 days as offset instead of 222
offset=310
# Find locale's abbreviated weekday name (e.g., Sun)
today=$(date -j +"%a")
# Check for Mon/Tue
if [[ "$today" == "Mon" ]] || [[ "$today" == "Tue" ]]; then
offset=$((offset+2))
fi
date_222_working_days_before_TodayYear=$(date -j -v-${offset}d +"%Y")
date_222_working_days_before_TodayMonth=$(date -j -v-${offset}d +"%m")
date_222_working_days_before_TodayDay=$(date -j -v-${offset}d +"%d")
And that should do it =)
应该这样做 =)
回答by Ashish K
You can get exact past date from the following in bash
您可以在 bash 中从以下内容中获得确切的过去日期
Number=222
current_date=$(date +%Y%m%d)
past_date=$(date -d "$current_date - $Number days" +%Y%m%d)
echo "$current_date\t$past_date"
Hope this helps !
希望这可以帮助 ！
回答by chrisaycock
date '+%Y' --date='222 days ago'
回答by djhaskin987
epoch=$(( `date '+%s'` - ( 24 * 60 * 60 * 222 ) ))
year=`date -d "@$epoch" '+%Y'`
month=`date -d "@$epoch" '+%m'`
day=`date -d "@$epoch" '+%d'`
Should do the trick.
应该做的伎俩。
回答by ThatComputerGuy
I would say easier solution would be
我会说更简单的解决方案是
222days_before_TodayYear = $(date -v -222d +%Y)
222days_before_TodayMonth = $(date -v -222d +%m)
222days_before_TodayDay = $(date -v -222d +%d)
