bash 我用于检查闰年的 shell 脚本显示错误
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时间：2020-09-18 13:30:00 来源:igfitidea点击:
My shell script for checking leap year is showing error
提问by Jens
#!/bin/bash
echo "Enter the year (YYYY)"
read year
if[ $((year % 4)) -eq 0]
then
if[ $((year % 100)) -eq 0]
then
if[ $((year % 400)) -eq 0]
then
echo "its a leap year"
else
echo "its not a leap year"
fi
else
echo "Its not a leap year"
fi
else
echo "its not a leap year"
fi
its showing error on 7th line and also on
它在第 7 行以及在第 7 行显示错误
[ $((year % 4)) -eq 0]
回答by anubhava
You've made it too complicated. Use this code to figure leap year:
你把它弄得太复杂了。使用此代码计算闰年：
isleap() {
year=
(( !(year % 4) && ( year % 100 || !(year % 400) ) )) &&
echo "leap year" || echo "not a leap"
}
Test it:
测试一下：
$ isleap 1900
not a leap
$ isleap 2000
leap year
$ isleap 2016
leap year
$ isleap 1800
not a leap
$ isleap 1600
leap year
回答by Jens
You miss some blanks which are nesessary in bash:
你错过了一些在 bash 中很必要的空格：
echo "Enter the year (YYYY)"
read year
if [ $((year % 4)) -eq 0 ]
then
if [ $((year % 100)) -eq 0 ]
then
if [ $((year % 400)) -eq 0 ]
then
echo "its a leap year"
else
echo "its not a leap year"
fi
else
echo "Its a leap year"
fi
else
echo "its not a leap year"
fi
回答by Prem Aswar
Test This Code:
测试此代码：
#!/bin/bash
echo "Enter a year to check - "
read y
year=$y
ans=`expr $year % 4`
if [ $ans -eq 0 ]
then
echo "Leap Year"
else
echo "Not a Leap Year"
fi
