bash - 在 bash 中，如何确定一组变量中是否只有一个有值？

Question

我有三个变量，我们称它们为 DATE、URI 和 URI_FILE。它们包含使用 getopt 处理命令行的结果:

while getopts ":d:p:u:hFA" opt; do

   case $opt in
       d) DATE=$OPTARG;;
       u) URI=${OPTARG};;
       U) URI_FILE=${OPTARG};;
       F) FORCE=yes;;
       A) ALL=yes;;

   esac

done

如果多个变量被设置为任何值(它们是备用命令行选项)，我需要失败并显示错误。我完全不知道如何优雅地测试它。我有一些看起来像的东西:

#!/bin/bash 

if [[ -z "${URI+x}" ]]
then
    URI_PRESENT=0
else
    URI_PRESENT=1
fi

if [[ -z "${URI_FILE+x}" ]]
then
    URI_FILE_PRESENT=0
else
    URI_FILE_PRESENT=1
fi

if [[ -z "${DATE+x}" ]]
then
    DATE_PRESENT=0
else
    DATE_PRESENT=1
fi


ARGCOUNT=$((${DATE_PRESENT} + ${URI_FILE_PRESENT} + ${URI_PRESENT}))

if (( ${ARGCOUNT} > 1 ))
then
    echo "Only one of date, uri and uri-file may be supplied"
    exit 1
fi

必须有一种更简洁的方法来做到这一点。建议？我将至少添加更多替代方案，这会很痛苦

Answer 1

来自 mickp's comment :

Why not have a single flag which gets incremented in every case expression and then test if it's greater than 1?

已实现:

while getopts ":d:p:u:hFA" opt; do
   case $opt in
   d) DATE=$OPTARG
      ((count++));;
   u) URI=${OPTARG}
      ((count++));;
   U) URI_FILE=${OPTARG}
      ((count++));;
   F) FORCE=yes;;
   A) ALL=yes;;
   esac
   if ((count > 1)); then
       echo "Only one of date, uri or uri-file may be supplied" >&2
       exit 1
   fi
done