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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 #!/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: |
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