shell 判断日期间隔及润年

    #!/bin/bash test.sh

  until

    echo "----------------------------------"

 echo "请输入您的选择:"

  echo "(1) 显示当前时间"

 echo "(2) 判断是否是润年"

   echo "(3) 时间间隔"

echo "(4) 退出"

 echo "----------------------------------"

 read input

    test $input = 4

    do

     case $input in

     1)

           t=` date '+%G-%m-%d %H:%M:%S' `

        echo $t

        ;;

      2)

        echo -n 请输入将要进行判断的年份:

        read year

         let "y1=$year % 4"

       let "y2=$year % 100"

       let "y3=$year % 400"

       if [ ! "$y1" -eq 0 ]

       then

         leap=0

       elif [ ! "$y2" -eq 0 ]

        then

       leap=1

        elif [ "$y3" -eq 0 ]

       then

        leap=1

       else

        leap=0

        fi

         if [ "$leap" -eq 1 ]

       then

        echo "$year 是闰年"

      else

         echo "$year 不是闰年"

             fi

        ;;

     3)

           read -p "请输入两个日期，格式(YYYYMMDD YYYYMMDD) :" start end

      ##将输入的日期转为的时间戳格式

      startDate=`date -d "${start}" +%s`

      endDate=`date -d "${end}" +%s`

      ##计算两个时间戳的差值除于每天86400s即为天数差

     stampDiff=`expr $endDate - $startDate`

     dayDiff=`expr $stampDiff / 86400`

     echo $dayDiff

         ;;

     4)echo " (1-4) "

 esac

     done