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#!/bin/bash
: '
方法一
LeapYear=`date +%Y`
if [ $[ $LeapYear % 400 ] -eq 0 ] ; then
echo "闰年"
elif [ $[ $LeapYear % 4 ] -eq 0 ] ; then
if [ $[ $LeapYear % 100 ] -ne 0 ] ; then
echo "闰年"
else
echo "不是闰年"
fi
else
echo "不是闰年"
fi
'
: '
# 方法二，利用布尔运算实现
LeapYear=`date +%Y`
if (( ( "$LeapYear" % 400 ) == "0" )) || (( ( "$LeapYear" % 4 == 0 ) && ( "$LeapYear" % 100 != "0" ) )) ; then
echo "闰年"
else
echo "不是闰年"
fi
# 这相当于let语句。如果你尝试类似$ [$ year％
'
# 利用read命令进行交互式判断实例
echo "请输入您想要判断是否为闰年的日期(四个数字),然后按enter：
read LeapYear
if ( ( ( " $LeapYear " % 400 ) = = 0 ) ) || ( ( ( " $LeapYear " % 4 = = 0 ) && ( " $LeapYear " % 100 != 0 ) ) ) ; then
echo "闰年"
else
echo "不是闰年"
fi
