在UNIX中处理日期 [英] Manipulate date in UNIX

问题描述

我想找到一种在UNIX中列出当前星期的所有日期的方法。
例如，如果今天是2018年7月17日，那么整周如何获取日期16-07、18-07、19-07等？即昨天，明天，后天等等。
由于未安装GNU，我无法在 date 命令中使用--date / d选项。其他任何命令或用户定义的函数来获取给定的所有日期（当前日期或任何日期）？

I want to find a way to list all the dates of the current week in UNIX. For example if today is 17-07-2018 then how to get dates 16-07, 18-07, 19-07 and so on for the whole week? I.e., yesterday, tomorrow, day after tomorrow etc. I cannot use --date/d options with date command as GNU is not installed. Any other commands or a user defined function to get all the dates of the given (current or any date)?

推荐答案

假设日期可以识别小时数的TZ偏移量，适用于当前日期：

Assuming your date recognizes TZ offsets in hours, this works for the current date:

#!/bin/sh

case $(date +%A) in
  (Monday)    off="  +0  -24 -48 -72 -96 -120 -144";;
  (Tuesday)   off=" +24   +0 -24 -48 -72  -96 -120";;
  (Wednesday) off=" +48  +24  +0 -24 -48  -72  -96";;
  (Thursday)  off=" +72  +48 +24  +0 -24  -48  -72";;
  (Friday)    off=" +96  +72 +48 +24  +0  -24  -48";;
  (Saturday)  off="+120  +96 +72 +48 +24   +0  -24";;
  (Sunday)    off="+144 +120 +96 +72 +48  +24   +0";;
esac
for o in $off; do
   TZ="$o" date +%d-%m
done

示例运行：

$ date
Sun Jul 15 23:13:43 CEST 2018
$ ./x.sh
09-07
10-07
11-07
12-07
13-07
14-07
15-07

leap年，下溢/上溢没有问题，因为它使用 date 的内置知识。

This has no problem with leap years, under/overflow, since it uses date's built-in knowledge.

PS：我刚刚发现我的 date （在FreeBSD上）采用UTC的偏移量，而不是系统时区的偏移量；您应该针对AIX进行调查，如果是这样，请通过将您当地时区的偏移量调整为UTC来相应地调整偏移量。

PS: I just discovered that my date (on FreeBSD) takes these offsets from UTC, not the systems time zone; you should investigate this for AIX and if so, adjust the offsets accordingly by your local time zone offset to UTC.

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