date命令，取得日期的前一天

lll0024

linux系统下可以显示：
#date --date ='a day ago'

但是，hp unix 11.系统怎样取得日期的前一天？

greendays

[root@Greendays /]# date
Tue Mar 24 17:19:27 CST 2009
[root@Greendays /]# date |awk '{print $1,$2,$3-1,$4,$5,$6}'
Tue Mar 23 17:19:29 CST 2009

飞鸿无痕

try  没玩过HP的
date -d 'yesterday'
或者按你说的
date --date='yesterday'

waker

http://bbs2.chinaunix.net/viewth ... p;page=1#pid4559524

我是DBA

不是这么简单的。还要考虑月头1号的情况。

magic6321

原帖由 greendays 于 2009-3-24 17:17 发表
[root@Greendays /]# date
Tue Mar 24 17:19:27 CST 2009
[root@Greendays /]# date |awk '{print $1,$2,$3-1,$4,$5,$6}'
Tue Mar 23 17:19:29 CST 2009

二楼真强.这也能想到.受教,受教了...

magic6321

如果是某月的一号??那....

lll0024

还要考虑月头1号的情况，
润年也要考虑。

我是DBA

麻烦看一下版版发的连接。。。

greendays

#!/bin/sh

# ydate: A Bourne shell script that
# prints yestarday's date
# Output Form: Month Day Year
# From Focus on Unix: http://unix.about.com

# Set the current month day and year.
month=`date +%m`
day=`date +%d`
year=`date +%Y`

# Add 0 to month. This is a
# trick to make month an unpadded integer.
month=`expr $month + 0`

# Subtract one from the current day.
day=`expr $day - 1`

# If the day is 0 then determine the last
# day of the previous month.
if [ $day -eq 0 ]; then
  
  # Find the preivous month.
  month=`expr $month - 1`  

  # If the month is 0 then it is Dec 31 of
  # the previous year.
  if [ $month -eq 0 ]; then
    month=12
    day=31
    year=`expr $year - 1`  

  # If the month is not zero we need to find
  # the last day of the month.
  else
    case $month in
      1|3|5|7|8|10|12) day=31;;
      4|6|9|11) day=30;;
      2)
        if [ `expr $year % 4` -eq 0 ]; then
          if [ `expr $year % 400` -eq 0 ]; then
            day=29
          elif [ `expr $year % 100` -eq 0 ]; then
            day=28
          else
            day=29
          fi
        else
          day=28
        fi
      ;;
    esac
  fi
fi

# Print the month day and year.
echo $month $day $year
exit 0