如何在bash中繞過以下參數的可選參數？

Question

TABLE=`echo "${1}" | tr '[:upper:]' '[:lower:]'` 
if [ $1 = -d ] 
    then TABLE=daminundation 
elif [ $1 = -b ] 
    then TABLE=burnscararea 
elif [ $1 = -r ] 
    then TABLE=riverpointinundation 
elif [ $1 = " " ] 
    then echo "User must input -d (daminundation), -b (burnscararea) 
    or -r (riverpointinundation)." 
fi 
SHAPEFILEPATH=${2} 
MERGEDFILENAME=${3} 
if [ -z $3 ] ; 
    then MERGEDFILENAME=merged.shp 
else 
    MERGEDFILENAME=${3} 
fi 
COLUMNNAME=${4} 
if [ -n $4 ] 
    then COLUMNNAME=$4 
fi 

$ 3 & $ 4是可選參數。但是，如果我選擇不使用$ 3，但我想使用$ 4，則它將以$ 3的形式讀取命令。對於其他方法感到困惑，我應該如何做到這一點，以避免下一個不需要的可選命令被繞過？

Answer 1

你可能想這樣的：

#!/bin/bash 

while getopts ":b :d :r" opt; do 
    case $opt in 
    b) 
     TABLE=burnscararea 
     ;; 
    d) 
     TABLE=daminundation 
     ;; 
    r) 
     TABLE=riverpointinundation 
     ;; 
    \?) 
     echo "Invalid option: -$OPTARG" >&2 
     exit 1 
     ;; 
    esac 
done 

shift $((OPTIND-1)) 

[ -z "$TABLE" ] && (echo "At least one of -b/-d/-r options must be provided"; exit 1;) 
[ $# -ne 3 ] && (echo "3 params expected!"; exit 1;) 
SHAPEFILEPATH="$2" 
MERGEDFILENAME="$3" 
COLUMNNAME="$4" 
# other stuff