判断输入的日期是否正确的shell脚本

今儿个讲得是判断输入的日期是否正确，有利用到我们之前03这个例子中的函数
下面是代码

#!/bin/sh
# valid-date -- Validates a date, taking into account leap year rules.

exceedsDaysInMonth()
{

 case $(echo $1|tr '[:upper:]' '[:lower:]') in
  jan* ) days=31  ;; feb* ) days=28  ;;
  mar* ) days=31  ;; apr* ) days=30  ;;
  may* ) days=31  ;; jun* ) days=30  ;;
  jul* ) days=31  ;; aug* ) days=31  ;;
  sep* ) days=30  ;; oct* ) days=31  ;;
  nov* ) days=30  ;; dec* ) days=31  ;;
  * ) echo "$0: Unknown month name $1" >&2; exit 1
  esac

  if [ $2 -lt 1 -o $2 -gt $days ] ; then
   return 1
  else
   return 0  # the day number is valid
  fi
}

isLeapYear()
{

 year=$1
 if [ "$((year % 4))" -ne 0 ] ; then
  return 1 # nope, not a leap year
 elif [ "$((year % 400))" -eq 0 ] ; then
  return 0 # yes, it's a leap year
 elif [ "$((year % 100))" -eq 0 ] ; then
  return 1
 else
  return 0
 fi
}
## Begin main script

if [ $# -ne 3 ] ; then
 echo "Usage: $0 month day year" >&2
 echo "Typical input fORMats are 8 3 2002" >&2
 exit 1
fi

# Normalize date and split back out returned values


if [ $? -eq 1 ] ; then
 exit 1    # error condition already reported by normdate
fi

monthnoToName()
{
 # Sets the variable 'month' to the appropriate value
 case $1 in
  01|1 ) monthd="Jan"  ;; 02|2 ) monthd="Feb"  ;;
  03|3 ) monthd="Mar"  ;; 04|4 ) monthd="Apr"  ;;
  05|5 ) monthd="May"  ;; 06|6 ) monthd="Jun"  ;;
  07|7 ) monthd="Jul"  ;; 08|8 ) monthd="Aug"  ;;
  09|9 ) monthd="Sep"  ;;   10) monthd="Oct"  ;;
    11) monthd="Nov"  ;;   12) monthd="Dec"  ;;
  * ) echo "$0: Unknown numeric month value $1" >&2; exit 1
  esac
  return 0
}

monthnoToName $1

month="$monthd"
 day="$2"
 year="$3"
 
if ! exceedsDaysInMonth $month "$2" ; then
 if [ "$month" = "Feb" -a "$2" -eq "29" ] ; then
  if ! isLeapYear $3 ; then
   echo "$0: $3 is not a leap year, so Feb doesn't have 29 days" >&2
   exit 1
  fi
 else
  echo "$0: bad day value: $month doesn't have $2 days" >&2
  exit 1
 fi
fi

echo "Valid date: $newdate"

exit 0

分析：
1）首先判断用户输入的参数个数是否正确，接着case $1 in 语句判断月份是否合理。
2）monthnoToName 函数之前出现在我们之前的第03个脚本案例中，用来转换输入的数字日期为字符串。
3） exceedsDaysInMonth 用来判断天数是否超过对应月的最大天数，紧跟着 if [ "$month" = "Feb" -a "$2" -eq "29" ] ; then if ! isLeapYear $3 ; then 用来判断闰年2月的特殊情况
4）总体的感觉是脚本还是很紧凑的，特别是在判断闰年与2月的关系的那段代码，有点意思。