YYYYMMDDHH在Unix中的shell脚本中进行日期操作 [英] YYYYMMDDHH date manipulation in shell script in Unix

问题描述

我的日期格式为YYYYMMDDHH，我想从中添加或减少小时数。 GNU date（date -d选项）在Linux中起着一种魅力的作用，但不像Solaris那样在非GNU env中执行。有人可以帮助我如何做到这一点吗？

I have a date in format "YYYYMMDDHH" and I want to add or subtract hours from it. The GNU date (date -d option) works like a charm in Linux but cant do this in non-GNU env like Solaris. Can someone please help me in how I can do this?

推荐答案

你应该检查 gdate 尚未安装在您的Solaris版本下（可能位于 / usr / gnu / bin / date ， / usr / sfw / bin / [g] date ， / usr / local / bin / [g] date ， / usr / csw / bin / [g] date 或只是 / usr / bin / gdate 取决于版本）。如果没有，应该很容易找到一个包含GNU日期并安装的软件包。

You should check if gdate is not already installed under your release of Solaris (might be in /usr/gnu/bin/date, /usr/sfw/bin/[g]date, /usr/local/bin/[g]date, /usr/csw/bin/[g]date or just /usr/bin/gdate depending on the version). If not, it should be easy to find a package containing GNU date and install it.

无论如何，这里是一个shell函数，它应该只是在Solaris版本的Solaris这样做我相信你想要的：

Anyway, here is a shell function that should just work under a stock Solaris release and that do what I believe you want:

f()
{
  echo $1 | perl -MTime::Local -nle '
  use POSIX 'strftime';
  $op='$2'*3600;
  $sec=timelocal(0,0,$4,$3,$2-1,$1) if /(\d{4})(\d{2})(\d{2})(\d{2})/;
  $sec=$sec+$op;
  print strftime "%Y%m%d%H\n", localtime($sec);'
}

---

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$ f 2014010112 -24
2013123112

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