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    判断输入的日期是否正确的shell脚本

    作者:admin 时间:2021-07-04 18:43

    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。

    jsjbwy