在 bash/unix 中获取上一个日期
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getting a previous date in bash/unix
提问by misguided
I am looking to get previous date in unix / shell script .
我希望在 unix/shell 脚本中获取上一个日期。
I am using the following code
我正在使用以下代码
date -d '1 day ago' +'%Y/%m/%d'
date -d '1 day ago' +'%Y/%m/%d'
But I am getting the following error.
但我收到以下错误。
date: illegal option -- d
date: illegal option -- d
As far as I've read on the inetrnet , it basically means I am using a older version of GNU. Can anyone please help with this.
就我在 inetrnet 上的阅读而言，这基本上意味着我使用的是旧版本的 GNU。任何人都可以帮忙解决这个问题。
Further Info
更多信息
unix> uname -a
unix> uname -a
SunOS Server 5.10 Generic_147440-19 sun4v sparc SUNW,Sun-Fire-T200
SunOS Server 5.10 Generic_147440-19 sun4v sparc SUNW,Sun-Fire-T200
Also The below command gives an error.
另外下面的命令给出了一个错误。
unix> date --version
unix> date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
采纳答案by devnull
Several solutions suggested here assume GNU coreutilsbeing present on the system. The following should work on Solaris:
此处建议的几种解决方案假定GNU coreutils存在于系统中。以下应该适用于 Solaris：
TZ=GMT+24 date +'%Y/%m/%d'
回答by kumarprd
try this:
尝试这个：
date --date="yesterday" +%Y/%m/%d
回答by Vladislav Ross
dtd="2015-06-19"
yesterday=$( date -d "${dtd} -1 days" +'%Y_%m_%d' )
echo $yesterday;
回答by Dan Pickard
you can use
您可以使用
date -d "30 days ago" +"%d/%m/%Y"
to get the date from 30 days ago, similarly you can replace 30 with x amount of days
要获取 30 天前的日期，同样您可以用 x 天数替换 30
回答by user3568717
In order to get 1 day back date using date command:
为了使用 date 命令获得 1 天的回溯日期：
date -v -1dIt will give (current date -1) means 1 day before .
date -v -1d它会给 (current date -1) 表示 1 天前。
date -v +1dThis will give (current date +1) means 1 day after.
date -v +1d这将给 (current date +1) 表示 1 天后。
Similarly below written code can be used in place of dto find out year,month etc
类似地，可以使用以下编写的代码代替d来找出年、月等
y-Year,
m-Month
w-Week
d-Day
H-Hour
M-Minute
S-Second
y-Year,
m-Month
w-Week
d-Day
H-Hour
M-Minute
S-Second
回答by Ravi Babu Mathi
the following script prints the previous date to the targetDate(specified Date or given date)
以下脚本将上一个日期打印到targetDate（指定日期或给定日期）
targetDate=2014-06-30
count=1
startDate=$( echo `date -d "${targetDate} -${count} days" +"%Y-%m-%d"`)
echo $startDate
回答by tripleee
SunOS ships with legacy BSD userland tools which often lack the expected modern options. See if you can get the XPG add-on (it's something like /usr/xpg4/bin/date) or install the GNU coreutilspackage if you can.
SunOS 附带了传统的 BSD 用户空间工具，这些工具通常缺乏预期的现代选项。看看您是否可以获得 XPG 附加组件（类似于/usr/xpg4/bin/date）或安装 GNUcoreutils软件包（如果可以）。
In the meantime, you might need to write your own simple date handling script. There are many examples on the net e.g. in Perl. E.g. this one:
同时，您可能需要编写自己的简单日期处理脚本。网上有很多例子，例如在 Perl 中。例如这个：
vnix$ perl -MPOSIX=strftime -le 'print strftime("%Y%m", localtime(time-86400))'
201304
(Slightly adapted, if you compare to the one behind the link.)
（稍微调整一下，如果与链接后面的比较的话。）
回答by misguided
I have used the following workaround to get to the required solution .
我已使用以下解决方法来获得所需的解决方案。
timeA=$(date +%Y%m)
sysD=$(date +%d)
print "Initial sysD $sysD">>LogPrint.log
sysD=$((sysD-1))
print "Final sysD $sysD">>LogPrint.log
finalTime=$timeA$sysD
I hope this is useful for people who are facing the same issue as me.
我希望这对与我面临同样问题的人有用。
回答by akash malbari
$ date '+%m/%d/%Y' --- current date
$ TZ=Etc/GMT+24 date '+%m/%d/%Y' -- one dayprevious date
Use time zone appropriately
适当使用时区
回答by venki
Try This: gdate -d "1 day ago" +"%Y/%m/%d"
试试这个： gdate -d "1 天前" +"%Y/%m/%d"
