Question

求Unix下shell脚本代码

用awk编写一个shell脚本，输入当前日期，获取到上个月当前日期。
本人新手求代码指点迷津，30分求导师

Answer 1

#!/bin/awk
BEGIN {
print "input a date:[yyyymmdd]" ;
getline str;
len=length(str);
if ( len !=8 )
{
printf ("error str=%s,len=%d" , str,len );
exit
}

new_day=get_last_mon_day(str);

printf( "new-day is :%s\n" , new_day );
}
function get_last_mon_day (str)
{
vYear=strtonum(substr( str , 1 , 4 ));
vMonth=strtonum(substr( str , 5 , 2 ));
vDay=strtonum(substr( str , 7 , 2 ));

if ( vMonth == 1 )
{
vMonth=12;
vYear -= 1;
}
else
{
vMonth -= 1 ;
}

isLY=is_leap_year(vYear);

max_day=get_mon_day( isLY , vMonth );
if ( vDay > max_day )
vDay=max_day ;

new_day=sprintf( "%04d%02d%02d" , vYear , vMonth , vDay );
return new_day ;
}
function is_leap_year( vYear )
{
m4=vYear%4 ;
m100=vYear%100;
m400=vYear%400;

if ( m4 == 0 && m100 != 0 )
return 1 ;
if ( m400 == 0 )
return 1;
return 0;
}

function get_mon_day( isLY , vMonth )
{
#awk has no switch statement!!
if ( isLY == 0 )
astr="31,28,31,30,31,30,31,31,30,31,30,31";
else
astr="31,29,31,30,31,30,31,31,30,31,30,31";
split(astr,am,",");
return am[vMonth];
}
我也在学习中，你先看看吧。把这些写到一个文件中，如：get_date.awk
然后在shell下执行 awk -f get_date.awk

Answer 2

$date;date -d '1 months ago'
Mon Sep 5 19:42:18 2011
Fri Aug 5 19:42:18 2011

Answer 3

其实你用命令date命令加几个参数就可以解决这个问题